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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

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

Answered by deleteduser1 last updated on 04/Nov/23

$$\frac{a^3}{4a+b+c}+\frac{a(4a+b+c)}{36}⩾2\sqrt{\frac{a^4}{36}}=\frac{a^2}{3}\left(true\right)$$$$Required:\mathrm{\Sigma }\frac{a(4a+b+c)}{36}⩽\mathrm{\Sigma }\frac{a^2}{6}$$$$\iff \frac{4(a^2+b^2+c^2)+2(ab+bc+ca)}{6}⩽(a^2+b^2+c^2)$$$$\iff a^2+b^2+c^2⩾ab+bc+ca\left(true\right)$$$$\Rightarrow \mathrm{\Sigma }\frac{a^3}{4a+b+c}⩾\mathrm{\Sigma }\left(\frac{a^2}{3}\right)-\mathrm{\Sigma }\frac{a(4a+b+c)}{36}⩾\mathrm{\Sigma }\frac{a^2}{3}-\mathrm{\Sigma }\frac{a^2}{6}$$$$=\mathrm{\Sigma }\frac{a^2}{6}◼$$

Commented by York12 last updated on 04/Nov/23

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

Commented by hardmath last updated on 05/Nov/23

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

Answered by York12 last updated on 04/Nov/23

$$\mathrm{WLOG}:a+b+c=3$$$$\Rightarrow \frac{a^3}{4a+b+c}+\frac{b^3}{4b+a+c}+\frac{c^3}{4c+a+b}=\frac{a^3}{3a+3}+\frac{b^3}{3b+3}+\frac{c^3}{3c+3}$$$$\mathrm{Assume}:\frac{a^3}{3a+3}⩾{ma}^2+n\iff a^3(1-3m)-3{ma}^2-3an-3n⩾0$$$$\mathrm{p}\left(\mathrm{a}\right)=a^3(1-3m)-3{ma}^2-3an-3n$$$${\mathrm{p}}^{\prime }\left(\mathrm{a}\right)=3a^2(1-3m)-6ma-3n$$$$p\left(1\right)=1-(6m+6n)=0\Rightarrow m+n=\frac{1}{6}$$$$p^{\prime }\left(1\right)=3-15\mathrm{m}-3\mathrm{n}\Rightarrow 5m+n=1$$$$\Rightarrow m=\frac{5}{24}andn=\frac{-1}{24}$$$$\Rightarrow \mathrm{p}\left(\mathrm{a}\right)=\frac{1}{24}\left({(a-1)}^2(3a+1)\right)⩾0\mathrm{\forall }a\in {\mathbb{R}}^{+}$$$$\Rightarrow \sum _{\mathrm{cyc}}\left(\frac{a^3}{4a+b+c}\right)⩾\frac{5(a^2+b^2+c^2)-3}{24}$$$$W\mathrm{e}\mathrm{have}{(a+b+c)}^2=9⩽3(a^2+b^2+c^2)$$$$\Rightarrow (a^2+b^2+c^2)⩾3$$$$\Rightarrow \sum _{\mathrm{cyc}}\left(\frac{a^3}{4a+b+c}\right)⩾\frac{5(a^2+b^2+c^2)-3}{24}⩾\frac{4(a^2+b^2+c^2)-3+3}{24}=\frac{a^2+b^2+c^2}{6}$$$$\left(H\mathrm{ence}\mathrm{proved}\right)$$$$$$

Commented by hardmath last updated on 05/Nov/23

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

Commented by York12 last updated on 05/Nov/23

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

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