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Question Number 55625 by gunawan last updated on 28/Feb/19

Commented by kaivan.ahmadi last updated on 28/Feb/19

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Commented by gunawan last updated on 28/Feb/19

$$\mathrm{Let}a,b,c\mathrm{be}\mathrm{positive}\mathrm{real}\mathrm{numbers}$$$$\mathrm{such}\mathrm{that}a+b+c=3.$$$$\mathrm{Prove}\mathrm{that}:$$$$\frac{a^6}{a^2+b}+\frac{b^6}{b^2+c}+\frac{c^6}{c^2+a}⩾\frac{3}{2}$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Mar/19

$$\frac{\mathit{A}+B}{2}⩾\sqrt{AB}⩾\frac{2AB}{A+B}$$$$\frac{1}{2}(\frac{a^2}{a^6}+\frac{b}{a^6})⩾\sqrt{\frac{a^{2}b}{a^6{a}^6}}⩾\frac{2\frac{a^2}{a^6}\times \frac{b}{a^6}}{\frac{a^2}{a^6}+\frac{b}{a^6}}$$$$(\frac{a^2}{a^6}+\frac{b}{a^6})⩾2\times \sqrt{\frac{b}{a^{10}}}$$$$\frac{a^2+b}{a^6}⩾\frac{2\sqrt{b}}{a^5}$$$$\frac{a^6}{a^2+b}⩽\frac{a^5}{2\sqrt{b}}$$$$givenexpressioni,ep⩽\frac{a^5}{2\sqrt{b}}+\frac{b^5}{2\sqrt{c}}+\frac{c^5}{2\sqrt{a}}$$$$letgivenexpression\frac{a^6}{a^2+b}+\frac{b^6}{b^2+c}+\frac{c^6}{c^2+a}=p$$$$\frac{p}{3}⩽\frac{1}{3}(\frac{a^5}{2\sqrt{b}}+\frac{b^5}{2\sqrt{c}}+\frac{c^5}{2\sqrt{a}})⩾{\left(\frac{a^5}{2\sqrt{b}}\times \frac{b^5}{2\sqrt{c}}\times \frac{c^5}{2\sqrt{a}}\right)}^{\frac{1}{3}}$$$$\frac{p}{3}⩽\frac{1}{3}(\frac{a^5}{2\sqrt{b}}+\frac{b^5}{2\sqrt{c}}+\frac{c^5}{2\sqrt{a}})⩾\frac{1}{2}{\left(abc\right)}^{\frac{9}{2}}$$$$\frac{p}{3}=\frac{1}{2}$$$$p=\frac{3}{2}$$$$$$$$$$$$since$$$$\frac{a+b+c}{3}⩾{\left(abc\right)}^{\frac{1}{3}}$$$$\frac{3}{3}⩾{\left(abc\right)}^{\frac{1}{3}}abc=1$$$$\frac{}{}$$

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