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

Answered by Berbere last updated on 08/Aug/24

$$a^2+b^2+\gamma ab\stackrel{AM-GM}{⩽}(\frac{\gamma }{2}+1)(a^2+b^2);\gamma >0$$$$\frac{a^3+b^3}{a^2+b^2+\gamma ab}⩾\frac{a+b}{\gamma +2}\Rightarrow \mathrm{\Sigma }\frac{a^3+b^3}{a^2+b^2+\gamma ab}⩾\frac{2(a+b+c)}{\gamma +2}\stackrel{AM-GM}{⩾}\frac{6}{\gamma +2}$$$$\frac{a^3+b^3}{2}⩾\frac{(a+b)(a^2+b^z)}{2};Tchebychev$$

Commented by hardmath last updated on 08/Aug/24

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

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