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Question Number 199923 by Rupesh123 last updated on 11/Nov/23

Answered by deleteduser1 last updated on 11/Nov/23

a+1+1≥3(a)^(1/3) ⇒Σ(a/(a+2))≤Σ(a^(2/3) /3) Power mean⇒(((a^2 +b^2 +c^2 )/3))^(1/2) ≥(((a^(2/3) +b^(2/3) +c^(2/3) )/3))^(3/2) ⇒Σa^(2/3) ≤3⇒Σ(a/(a+2))≤(3/3)=1

$${a}+\mathrm{1}+\mathrm{1}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{a}}\Rightarrow\Sigma\frac{{a}}{{a}+\mathrm{2}}\leqslant\Sigma\frac{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{3}} \\ $$$${Power}\:{mean}\Rightarrow\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\geqslant\left(\frac{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}} +{c}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow\Sigma{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \leqslant\mathrm{3}\Rightarrow\Sigma\frac{{a}}{{a}+\mathrm{2}}\leqslant\frac{\mathrm{3}}{\mathrm{3}}=\mathrm{1} \\ $$

Commented by Rupesh123 last updated on 11/Nov/23

Very elegant, sir!

Answered by witcher3 last updated on 12/Nov/23

⇔3−2((1/(a+2))+(1/(b+2))+(1/(c+2)))≤1 Σ_(Cycl) (1/(a+2))≥(((1+1+1)^2 )/(6+a+b+c))=(9/(6_ +a+b+c)) Ar−Quadratic Mean cauchy shwartz ⇒∣1.a+1.b+1.c∣≤(√(3.(a^2 +b^2 +c^2 )))=3 ⇔6+a+b+c≤9 ⇔(9/(6+a+b+c))≥1 ⇒Σ_(cycl) (1/(a+2))≤1 ⇔3−2(Σ_(cycl) (1/(a+2)))=Σ(a/(a+2))≤3−2=1

$$\Leftrightarrow\mathrm{3}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{a}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{b}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{c}+\mathrm{2}}\right)\leqslant\mathrm{1} \\ $$$$\underset{\mathrm{Cycl}} {\sum}\frac{\mathrm{1}}{\mathrm{a}+\mathrm{2}}\geqslant\frac{\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{6}+\mathrm{a}+\mathrm{b}+\mathrm{c}}=\frac{\mathrm{9}}{\mathrm{6}_{} +\mathrm{a}+\mathrm{b}+\mathrm{c}} \\ $$$$\mathrm{Ar}−\mathrm{Quadratic}\:\mathrm{Mean} \\ $$$$\mathrm{cauchy}\:\mathrm{shwartz}\: \\ $$$$\Rightarrow\mid\mathrm{1}.\mathrm{a}+\mathrm{1}.\mathrm{b}+\mathrm{1}.\mathrm{c}\mid\leqslant\sqrt{\mathrm{3}.\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)}=\mathrm{3} \\ $$$$\Leftrightarrow\mathrm{6}+\mathrm{a}+\mathrm{b}+\mathrm{c}\leqslant\mathrm{9} \\ $$$$\Leftrightarrow\frac{\mathrm{9}}{\mathrm{6}+\mathrm{a}+\mathrm{b}+\mathrm{c}}\geqslant\mathrm{1} \\ $$$$\Rightarrow\underset{\mathrm{cycl}} {\sum}\frac{\mathrm{1}}{\mathrm{a}+\mathrm{2}}\leqslant\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{3}−\mathrm{2}\left(\underset{\mathrm{cycl}} {\sum}\frac{\mathrm{1}}{\mathrm{a}+\mathrm{2}}\right)=\Sigma\frac{\mathrm{a}}{\mathrm{a}+\mathrm{2}}\leqslant\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$$ \\ $$

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