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Question Number 210355 by hardmath last updated on 08/Aug/24

Answered by Berbere last updated on 08/Aug/24

a^2 +b^2 +γab≤^(AM−GM) ((γ/2)+1)(a^2 +b^2 );γ>0 ((a^3 +b^3 )/(a^2 +b^2 +γab))≥((a+b)/(γ+2))⇒Σ((a^3 +b^3 )/(a^2 +b^2 +γab))≥((2(a+b+c))/(γ+2))≥^(AM−GM) (6/(γ+2)) ((a^3 +b^3 )/2)≥(((a+b)(a^2 +b^z ))/2);Tchebychev

$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\gamma{ab}\overset{{AM}−{GM}} {\leqslant}\left(\frac{\gamma}{\mathrm{2}}+\mathrm{1}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right);\gamma>\mathrm{0} \\ $$$$\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\gamma{ab}}\geqslant\frac{{a}+{b}}{\gamma+\mathrm{2}}\Rightarrow\Sigma\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\gamma{ab}}\geqslant\frac{\mathrm{2}\left({a}+{b}+{c}\right)}{\gamma+\mathrm{2}}\overset{{AM}−{GM}} {\geqslant}\frac{\mathrm{6}}{\gamma+\mathrm{2}} \\ $$$$\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{\mathrm{2}}\geqslant\frac{\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{{z}} \right)}{\mathrm{2}};{Tchebychev} \\ $$

Commented by hardmath last updated on 08/Aug/24

thank you dear professor cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$

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