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Question Number 220190    Answers: 1   Comments: 0

i^i =e^(−(π/2)) and we can renote complex number i as ((0,(−1)),(1,( 0)) ) i^i = ((0,(−1)),(1,( 0)) )^ ((0,(−1)),(1,( 0)) ) But why Matrix Exponent Calculate Dosen′t defined?? I mean A,B∈mat(m,m) why A^B dosen′t defined??

$${i}^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} \: \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{renote}\:\mathrm{complex}\:\mathrm{number}\:\boldsymbol{{i}}\:\mathrm{as}\:\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\boldsymbol{{i}}^{\boldsymbol{{i}}} =\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix}^{\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix}} \: \\ $$$$\:\mathrm{But}\:\mathrm{why}\:\mathrm{Matrix}\:\mathrm{Exponent}\:\mathrm{Calculate}\:\mathrm{Dosen}'\mathrm{t}\:\mathrm{defined}?? \\ $$$$\:\mathrm{I}\:\mathrm{mean}\:{A},{B}\in\mathrm{mat}\left({m},{m}\right) \\ $$$$\mathrm{why}\:\mathrm{A}^{\mathrm{B}} \:\mathrm{dosen}'\mathrm{t}\:\mathrm{defined}?? \\ $$

Question Number 220184    Answers: 2   Comments: 0

∫_0 ^(π/(12)) (√((sec^4 α+5sec^5 αsin α)/((2−sec^2 α)(125tan^3 α+25tan^2 α+5tan α+1))))dα

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \sqrt{\frac{\mathrm{sec}^{\mathrm{4}} \alpha+\mathrm{5sec}^{\mathrm{5}} \alpha\mathrm{sin}\:\alpha}{\left(\mathrm{2}−\mathrm{sec}^{\mathrm{2}} \alpha\right)\left(\mathrm{125tan}^{\mathrm{3}} \alpha+\mathrm{25tan}^{\mathrm{2}} \alpha+\mathrm{5tan}\:\alpha+\mathrm{1}\right)}}{d}\alpha \\ $$

Question Number 220323    Answers: 0   Comments: 0

Q1. 𝛀;={(x,y);x^2 +y^2 ≤1} (1/2) ∮_( ∂𝛀) x∙dy−y∙dy=?? Q2. S; R^2 →R^3 S(u,v)=rsin(u)cos(v)e_1 ^→ +rsin(u)sin(v)e_2 ^→ +rcos(u)e_3 ^→ F^→ ;R^3 →R^3 F^→ (x,y,z)=−(x/( (√(x^2 +y^2 +z^2 ))))e_1 ^→ −(y/( (√(x^2 +y^2 +z^2 ))))e_2 ^→ −(z/( (√(x^2 +y^2 +z^2 ))))e_3 ^→ ∫∫_( D) F^→ ∙dS^→ =∫∫∫_( K) ▽^→ ∙F^→ dV=??? Q3. if ∮_( C) F^→ ∙dl=0 Prove ▽^→ ×F^→ =0 Q4. Prove if ▽^→ ×F^→ ≠0 → ∮_( C) F^→ ∙dl≠0

$$\mathrm{Q1}.\:\boldsymbol{\Omega};=\left\{\left({x},{y}\right);{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leq\mathrm{1}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\oint_{\:\partial\boldsymbol{\Omega}} \:{x}\centerdot\mathrm{d}{y}−{y}\centerdot\mathrm{d}{y}=?? \\ $$$$\mathrm{Q2}.\:\boldsymbol{\mathcal{S}};\:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}^{\mathrm{3}} \\ $$$$\boldsymbol{\mathcal{S}}\left({u},{v}\right)={r}\mathrm{sin}\left({u}\right)\mathrm{cos}\left({v}\right)\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} +{r}\mathrm{sin}\left({u}\right)\mathrm{sin}\left({v}\right)\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} +{r}\mathrm{cos}\left({u}\right)\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}};\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}^{\mathrm{3}} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)=−\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −\frac{{z}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\:\int\int_{\:\mathcal{D}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}=\int\int\int_{\:\boldsymbol{{K}}} \:\overset{\rightarrow} {\bigtriangledown}\centerdot\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\:\mathrm{d}{V}=??? \\ $$$$\mathrm{Q3}. \\ $$$$\mathrm{if}\:\:\oint_{\:\mathcal{C}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\boldsymbol{\mathrm{l}}=\mathrm{0}\:\:\mathrm{Prove}\:\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}=\mathrm{0} \\ $$$$\mathrm{Q4}. \\ $$$$\mathrm{Prove}\:\:\mathrm{if}\:\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\neq\mathrm{0}\:\rightarrow\:\oint_{\:\mathcal{C}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\boldsymbol{\mathrm{l}}\neq\mathrm{0} \\ $$

Question Number 220179    Answers: 1   Comments: 0

∫ (ds/( (√(s^2 +1))(s+(√(s^2 +1)))^(−ν) ))

$$\int\:\:\:\frac{\mathrm{d}{s}}{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} } \\ $$

Question Number 220178    Answers: 1   Comments: 0

evaluate ∫_w ^( ∞) ((e^s ∙𝚪(0,s))/s)ds

$$\mathrm{evaluate} \\ $$$$\int_{{w}} ^{\:\infty} \:\frac{{e}^{{s}} \centerdot\boldsymbol{\Gamma}\left(\mathrm{0},{s}\right)}{{s}}\mathrm{d}{s} \\ $$

Question Number 220177    Answers: 1   Comments: 0

∫_(−1) ^( 0) cos(((ln(z+1))/z)) dz

$$\int_{−\mathrm{1}} ^{\:\mathrm{0}} \:\:\mathrm{cos}\left(\frac{\mathrm{ln}\left({z}+\mathrm{1}\right)}{{z}}\right)\:\mathrm{d}{z} \\ $$

Question Number 220176    Answers: 0   Comments: 0

∫ (dx/(1 + x^7 ))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{7}} }\: \\ $$$$ \\ $$

Question Number 220164    Answers: 2   Comments: 0

if α^2 −5α+2=0 & β^2 −5β+2=0 then ((4α+β^5 )/(5β^2 ))=?

$${if}\:\:\alpha^{\mathrm{2}} −\mathrm{5}\alpha+\mathrm{2}=\mathrm{0}\:\:\&\:\:\beta^{\mathrm{2}} −\mathrm{5}\beta+\mathrm{2}=\mathrm{0} \\ $$$${then}\:\:\frac{\mathrm{4}\alpha+\beta^{\mathrm{5}} }{\mathrm{5}\beta^{\mathrm{2}} }=? \\ $$

Question Number 220160    Answers: 0   Comments: 0

for all x , y [0 , 1] ; prove that; [ (((x^3 + y^3 + 𝛇(3)))^(1/(3 )) /(1 + e^(−x^2 y^2 ) )) + (((x^4 + 𝚪(y+1)))^(1/(4 )) /((1 + y^2 )^(1/3) )) + ((ln(1 + x^5 + y^5 ))/( (√(1 + x^2 + y^2 )))) + Li_2 (xy) + ((√(x^6 + y^6 +1 ))/((1 + x^3 y^3 )^(1/2) )) ≤ (e^(xy) /(1 + x + y )) + ((ln (1 + x^2 + y^2 ) ))^(1/(3 )) + ((2𝛇(2))/( (√(1 + x^2 y^2 )))) + ((x^8 + y^8 + 1))^(1/(4 )) ]

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x}\:,\:{y}\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\left[\:\frac{\sqrt[{\mathrm{3}\:\:}]{\boldsymbol{{x}}^{\mathrm{3}} \:+\:\boldsymbol{{y}}^{\mathrm{3}} \:+\:\boldsymbol{\zeta}\left(\mathrm{3}\right)}}{\mathrm{1}\:+\:\boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} } \:}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{x}}^{\mathrm{4}} +\:\boldsymbol{\Gamma}\left(\boldsymbol{{y}}+\mathrm{1}\right)}}{\left(\mathrm{1}\:+\:\boldsymbol{{y}}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} }\:+\:\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{5}} \:+\:\boldsymbol{{y}}^{\mathrm{5}} \right)}{\:\sqrt{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\boldsymbol{{xy}}\right)\:+\:\frac{\sqrt{\boldsymbol{{x}}^{\mathrm{6}} \:+\:\boldsymbol{{y}}^{\mathrm{6}} \:+\mathrm{1}\:}}{\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{2}} }\: \\ $$$$\left.\:\:\:\:\:\:\leqslant\:\frac{\boldsymbol{{e}}^{\boldsymbol{{xy}}} }{\mathrm{1}\:+\:\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:}\:+\:\sqrt[{\mathrm{3}\:\:}]{\boldsymbol{\mathrm{ln}}\:\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}} \right)\:}\:+\:\frac{\mathrm{2}\boldsymbol{\zeta}\left(\mathrm{2}\right)}{\:\sqrt{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} }}\:+\:\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{x}}^{\mathrm{8}} \:+\:\boldsymbol{{y}}^{\mathrm{8}} \:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\right]\:\: \\ $$$$ \\ $$$$\:\: \\ $$

Question Number 220159    Answers: 0   Comments: 1

for all x, y ∈ [0 , 1] ; prove that; (1/( (√(1 + x^4 )))) + (2/( (√(1 + y^4 )))) + (2/( (√(4 + (x + y)^4 )))) + ((2(√2))/( (√(2+ x^2 y^2 + y^3 )))) ≤ (2/( (√(1 + x^2 y^2 )))) + (2/(^4 (√(1 + x^5 + y^5 )))) + ln(e+((x^3 y+y^3 x)/(1 + xy))) + (1/((1+x+y)^3 ))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x},\:{y}\:\in\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}\:+\:\left({x}\:+\:{y}\right)^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:{y}^{\mathrm{3}} }}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\leqslant\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} }}\:+\:\frac{\mathrm{2}}{\:^{\mathrm{4}} \sqrt{\mathrm{1}\:+\:{x}^{\mathrm{5}} \:+\:{y}^{\mathrm{5}} }}\:+\:\mathrm{ln}\left({e}+\frac{{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {x}}{\mathrm{1}\:+\:{xy}}\right)\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}+{y}\right)^{\mathrm{3}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 220141    Answers: 0   Comments: 0

Σ_(n=1) ^∞ Σ_(m=1) ^∞ (1/((n^2 +m^2 )^(3/2) ))=?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{m}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }=? \\ $$

Question Number 220131    Answers: 2   Comments: 0

If f(x,y)=(((x^2 +y^2 )^n )/(2n(2n−1)))+xφ((y/x))+Ψ((y/x)), then using Euler′s theorem on homogenous functions,show that x^2 ((δ^2 f)/(δx^2 ))+2xy((δ^2 f)/(δxδy))+y^2 ((δ^2 f)/(δy^2 ))=(x^2 +y^2 )^n

$${If}\:\:\:{f}\left({x},{y}\right)=\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{{n}} }{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}+{x}\phi\left(\frac{{y}}{{x}}\right)+\Psi\left(\frac{{y}}{{x}}\right), \\ $$$${then}\:{using}\:{Euler}'{s}\:{theorem}\:{on}\:{homogenous}\:{functions},{show}\:{that} \\ $$$${x}^{\mathrm{2}} \frac{\delta^{\mathrm{2}} {f}}{\delta{x}^{\mathrm{2}} }+\mathrm{2}{xy}\frac{\delta^{\mathrm{2}} {f}}{\delta{x}\delta{y}}+{y}^{\mathrm{2}} \frac{\delta^{\mathrm{2}} {f}}{\delta{y}^{\mathrm{2}} }=\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{{n}} \\ $$

Question Number 220121    Answers: 1   Comments: 0

Let H_h =p_(h+1) /p_h , p_h ∈P , p_1 =2 Π_(h=1) ^∞ H_h =?? (Π_(h=1) ^∞ H_h =(3/2)∙(5/3)∙(7/5).........)

$$\mathrm{Let}\:{H}_{{h}} ={p}_{{h}+\mathrm{1}} /{p}_{{h}} \:,\:{p}_{{h}} \in\mathbb{P}\:,\:{p}_{\mathrm{1}} =\mathrm{2} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\prod}}\:{H}_{{h}} =??\:\left(\underset{{h}=\mathrm{1}} {\overset{\infty} {\prod}}\:{H}_{{h}} =\frac{\mathrm{3}}{\mathrm{2}}\centerdot\frac{\mathrm{5}}{\mathrm{3}}\centerdot\frac{\mathrm{7}}{\mathrm{5}}.........\right) \\ $$

Question Number 220321    Answers: 1   Comments: 2

Question Number 220115    Answers: 3   Comments: 0

Question Number 220114    Answers: 5   Comments: 0

Question Number 220113    Answers: 1   Comments: 0

Question Number 220111    Answers: 2   Comments: 1

Question Number 220110    Answers: 2   Comments: 0

Question Number 220108    Answers: 7   Comments: 0

Question Number 220104    Answers: 1   Comments: 0

Question Number 220103    Answers: 8   Comments: 0

Question Number 220101    Answers: 0   Comments: 0

Question Number 220097    Answers: 1   Comments: 0

x ∈ Q ; x ≠ 1 (7/(x − 1)) + (6/x) − (4/(x + 1)) + ((3x + 5)/(x^2 − 1)) = (1/x)

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\in\:\mathbb{Q}\:\:\:;\:\:\:\:{x}\:\neq\:\mathrm{1} \\ $$$$\:\frac{\mathrm{7}}{{x}\:−\:\mathrm{1}}\:+\:\frac{\mathrm{6}}{{x}}\:−\:\frac{\mathrm{4}}{{x}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{3}{x}\:+\:\mathrm{5}}{{x}^{\mathrm{2}} \:−\:\mathrm{1}}\:=\:\frac{\mathrm{1}}{{x}} \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 220096    Answers: 1   Comments: 0

let a, b, c, d, e is a positive real numbers and K = a + b + c + d + e +1 . prove that; Σ_(cyc) (1/(k−a)) < (1/4) ((((e^3 d^3 c))^(1/(4 )) /(c^(3/4) d^(1/2) e^(1/4) (√a))) + (((d^( 3) c^2 b))^(1/(4 )) /(d^( 3/4) c^(1/2) b^(1/4) (√e))) + (((c^3 b^2 a))^(1/(4 )) /(c^(3/4) b^(1/2) a^(1/4) (√d))) + (((b^3 a^2 e))^(1/(4 )) /(b^(3/4) a^(1/2) e^(1/4) (√c))) + (((a^3 e^2 d))^(1/(4 )) /(a^(3/4) e^(1/2) d^(1/4) (√b))))

$$ \\ $$$$\:\:\:\mathrm{let}\:{a},\:{b},\:{c},\:{d},\:{e}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{K}\:=\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:+\:{e}\:+\mathrm{1}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\underset{\boldsymbol{{cyc}}} {\sum}\:\frac{\mathrm{1}}{\boldsymbol{{k}}−\boldsymbol{{a}}}\:<\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(\frac{\sqrt[{\mathrm{4}\:\:\:\:}]{\boldsymbol{{e}}^{\mathrm{3}} \boldsymbol{{d}}^{\mathrm{3}} \boldsymbol{{c}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{a}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:\:}]{\boldsymbol{{d}}^{\:\mathrm{3}} \boldsymbol{{c}}^{\mathrm{2}} \boldsymbol{{b}}}}{\boldsymbol{{d}}^{\:\mathrm{3}/\mathrm{4}} \boldsymbol{{c}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{b}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{e}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{c}}^{\mathrm{3}} \boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{a}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{b}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{a}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{d}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{b}}^{\mathrm{3}} \boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{e}}}}{\boldsymbol{{b}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{a}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{c}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{a}}^{\mathrm{3}} \boldsymbol{{e}}^{\mathrm{2}} \boldsymbol{{d}}}}{\boldsymbol{{a}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{b}}}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\: \\ $$

Question Number 220094    Answers: 1   Comments: 0

let n ≥ 2 ∈ Z and x_1 , x_2 , ..., x_n are a positive real numbers such that Σ_(i=1) ^n x_i = n , prove that Σ_(i=1) ^n (x_i ^n /(x_1 + ∙∙∙ + x_i ^ + ∙∙∙ + x_n )) ≥ (n/(n − 1))

$$ \\ $$$$\:\:\:\:\mathrm{let}\:{n}\:\geqslant\:\mathrm{2}\:\in\:\mathbb{Z}\:\mathrm{and}\:{x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:...,\:{x}_{{n}} \:\mathrm{are}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{such}\:\mathrm{that}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:{x}_{{i}} \:=\:{n}\:,\:\mathrm{prove}\:\mathrm{that}\:\:\:\: \\ $$$$\:\:\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{x}_{{i}} ^{{n}} }{{x}_{\mathrm{1}} +\:\centerdot\centerdot\centerdot\:+\:\hat {{x}}_{{i}} \:+\:\centerdot\centerdot\centerdot\:+\:{x}_{{n}} }\:\geqslant\:\frac{{n}}{{n}\:−\:\mathrm{1}} \\ $$$$ \\ $$

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