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Question Number 199509 by hardmath last updated on 04/Nov/23

Commented by York12 last updated on 04/Nov/23

Commented by York12 last updated on 04/Nov/23

That is more general

$$\mathrm{That}\:\mathrm{is}\:\mathrm{more}\:\mathrm{general}\: \\ $$

Answered by deleteduser1 last updated on 04/Nov/23

(a^3 /(4a+b+c))+((a(4a+b+c))/(36))≥2(√(a^4 /(36)))=(a^2 /3)(true) Required: Σ((a(4a+b+c))/(36))≤Σ(a^2 /6) ⇔((4(a^2 +b^2 +c^2 )+2(ab+bc+ca))/6)≤(a^2 +b^2 +c^2 ) ⇔a^2 +b^2 +c^2 ≥ab+bc+ca(true) ⇒Σ(a^3 /(4a+b+c))≥Σ((a^2 /3))−Σ((a(4a+b+c))/(36))≥Σ(a^2 /3)−Σ(a^2 /6) =Σ(a^2 /6) ■

$$\frac{{a}^{\mathrm{3}} }{\mathrm{4}{a}+{b}+{c}}+\frac{{a}\left(\mathrm{4}{a}+{b}+{c}\right)}{\mathrm{36}}\geqslant\mathrm{2}\sqrt{\frac{{a}^{\mathrm{4}} }{\mathrm{36}}}=\frac{{a}^{\mathrm{2}} }{\mathrm{3}}\left({true}\right) \\ $$$${Required}:\:\Sigma\frac{{a}\left(\mathrm{4}{a}+{b}+{c}\right)}{\mathrm{36}}\leqslant\Sigma\frac{{a}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Leftrightarrow\frac{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\mathrm{2}\left({ab}+{bc}+{ca}\right)}{\mathrm{6}}\leqslant\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ab}+{bc}+{ca}\left({true}\right) \\ $$$$\Rightarrow\Sigma\frac{{a}^{\mathrm{3}} }{\mathrm{4}{a}+{b}+{c}}\geqslant\Sigma\left(\frac{{a}^{\mathrm{2}} }{\mathrm{3}}\right)−\Sigma\frac{{a}\left(\mathrm{4}{a}+{b}+{c}\right)}{\mathrm{36}}\geqslant\Sigma\frac{{a}^{\mathrm{2}} }{\mathrm{3}}−\Sigma\frac{{a}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$=\Sigma\frac{{a}^{\mathrm{2}} }{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$

Commented by York12 last updated on 04/Nov/23

Good JOB !

$$\mathrm{Good}\:\mathrm{JOB}\:! \\ $$

Commented by hardmath last updated on 05/Nov/23

thank you ser

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{ser} \\ $$

Answered by York12 last updated on 04/Nov/23

WLOG :a+b+c=3 ⇒(a^3 /(4a+b+c))+(b^3 /(4b+a+c))+(c^3 /(4c+a+b))=(a^3 /(3a+3))+(b^3 /(3b+3))+(c^3 /(3c+3)) Assume: (a^3 /(3a+3))≥ma^2 +n ⇔a^3 (1−3m)−3ma^2 −3an−3n≥0 p(a)=a^3 (1−3m)−3ma^2 −3an−3n p′(a)=3a^2 (1−3m)−6ma−3n p(1)=1−(6m+6n)=0 ⇒m+n=(1/6) p′(1)=3−15m−3n⇒5m+n=1 ⇒m=(5/(24)) and n=((−1)/(24)) ⇒p(a)=(1/(24))((a−1)^2 (3a+1))≥0 ∀ a ∈ R^+ ⇒Σ_(cyc) ((a^3 /(4a+b+c)))≥((5(a^2 +b^2 +c^2 )−3)/(24)) We have (a+b+c)^2 =9≤3(a^2 +b^2 +c^2 ) ⇒(a^2 +b^2 +c^2 )≥3 ⇒Σ_(cyc) ((a^3 /(4a+b+c)))≥((5(a^2 +b^2 +c^2 )−3)/(24))≥((4(a^2 +b^2 +c^2 )−3+3)/(24))=((a^2 +b^2 +c^2 )/6) (Hence proved)

$$\mathrm{WLOG}\::{a}+{b}+{c}=\mathrm{3}\: \\ $$$$\Rightarrow\frac{{a}^{\mathrm{3}} }{\mathrm{4}{a}+{b}+{c}}+\frac{{b}^{\mathrm{3}} }{\mathrm{4}{b}+{a}+{c}}+\frac{{c}^{\mathrm{3}} }{\mathrm{4}{c}+{a}+{b}}=\frac{{a}^{\mathrm{3}} }{\mathrm{3}{a}+\mathrm{3}}+\frac{{b}^{\mathrm{3}} }{\mathrm{3}{b}+\mathrm{3}}+\frac{{c}^{\mathrm{3}} }{\mathrm{3}{c}+\mathrm{3}} \\ $$$$\mathrm{Assume}:\:\frac{{a}^{\mathrm{3}} }{\mathrm{3}{a}+\mathrm{3}}\geqslant{ma}^{\mathrm{2}} +{n}\:\Leftrightarrow{a}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{3}{m}\right)−\mathrm{3}{ma}^{\mathrm{2}} −\mathrm{3}{an}−\mathrm{3}{n}\geqslant\mathrm{0} \\ $$$$\mathrm{p}\left(\mathrm{a}\right)={a}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{3}{m}\right)−\mathrm{3}{ma}^{\mathrm{2}} −\mathrm{3}{an}−\mathrm{3}{n} \\ $$$$\mathrm{p}'\left(\mathrm{a}\right)=\mathrm{3}{a}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{3}{m}\right)−\mathrm{6}{ma}−\mathrm{3}{n} \\ $$$${p}\left(\mathrm{1}\right)=\mathrm{1}−\left(\mathrm{6}{m}+\mathrm{6}{n}\right)=\mathrm{0}\:\Rightarrow{m}+{n}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${p}'\left(\mathrm{1}\right)=\mathrm{3}−\mathrm{15m}−\mathrm{3n}\Rightarrow\mathrm{5}{m}+{n}=\mathrm{1} \\ $$$$\Rightarrow{m}=\frac{\mathrm{5}}{\mathrm{24}}\:{and}\:\:{n}=\frac{−\mathrm{1}}{\mathrm{24}}\: \\ $$$$\Rightarrow\mathrm{p}\left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{24}}\left(\left({a}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{3}{a}+\mathrm{1}\right)\right)\geqslant\mathrm{0}\:\:\forall\:{a}\:\in\:\mathbb{R}^{+} \\ $$$$\Rightarrow\underset{\mathrm{cyc}} {\sum}\left(\frac{{a}^{\mathrm{3}} }{\mathrm{4}{a}+{b}+{c}}\right)\geqslant\frac{\mathrm{5}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{3}}{\mathrm{24}} \\ $$$${W}\mathrm{e}\:\mathrm{have}\:\left({a}+{b}+{c}\right)^{\mathrm{2}} =\mathrm{9}\leqslant\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\geqslant\mathrm{3} \\ $$$$\Rightarrow\underset{\mathrm{cyc}} {\sum}\left(\frac{{a}^{\mathrm{3}} }{\mathrm{4}{a}+{b}+{c}}\right)\geqslant\frac{\mathrm{5}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{3}}{\mathrm{24}}\geqslant\frac{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{3}+\mathrm{3}}{\mathrm{24}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\left({H}\mathrm{ence}\:\mathrm{proved}\right) \\ $$$$ \\ $$

Commented by hardmath last updated on 05/Nov/23

thank you ser

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{ser} \\ $$

Commented by York12 last updated on 05/Nov/23

ur welcome

$$\mathrm{ur}\:\mathrm{welcome} \\ $$

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