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Let $a,b,c>0$ such that $ab+bc+ca>0$. Prove the inequality $$\frac{a^3+abc}{b^2+c^2}+\frac{b^3+abc}{c^2+a^2}+\frac{c^3+abc}{a^2+b^2}\ge a+b+c$$


My try 1: S.O.S:

$$​\Leftrightarrow \frac{a^3+abc}{b^2+c^2}-a+\frac{b^3+abc}{c^2+a^2}-b+\frac{c^3+abc}{a^2+b^2}-c\ge 0$$

$$\Leftrightarrow Σ_{cyc}\left(\frac{a\left(a-b\right)\left(a+b\right)+ac\left(b-c\right)}{b^2+c^2}\right)\ge 0$$

$$\Leftrightarrow Σ_{cyc}\left(\left(a-b\right)\left(\frac{a^4+a^3b+a^2b^2+ab^3+b^3c+bc^3}{\left(b^2+c^2\right)\left(a^2+b^2\right)}\right)\right)\ge 0$$

I don't know how about analyze it

My try 2: I assume that: $a^2+b^2+c^2=3$. Have : $a+b+c\le \sqrt{3\left(a^2+b^2+c^2\right)}=3$

And we need to prove:$$\frac{a^3+abc}{b^2+c^2}+\frac{b^3+abc}{c^2+a^2}+\frac{c^3+abc}{a^2+b^2}\ge 3$$

$$Σ_{cyc}\frac{a^3}{3-a^2}+abc\left(Σ_{cyc}\frac{1}{a^2+b^2}\right)\ge 3$$

I solved $Σ_{cyc}\frac{a^3}{3-a^2}\ge \frac{3}{2}$ but $abc\left(Σ_{cyc}\frac{1}{a^2+b^2}\right)\ge \frac{3}{2}$ i have no idea about solve $abc$, it make me hard prove it.

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I think your way is not so good.

Indeed, we need to prove that $$\sum_{cyc}\left(\frac{a^3+abc}{b^2+c^2}-a\right)\geq0$$ or $$\sum_{cyc}\frac{a(a^2+bc-b^2-c^2)}{b^2+c^2}\geq0$$ or $$\sum_{cyc}\frac{a((a-b)(a+2b-c)-(c-a)(a+2c-b))}{b^2+c^2}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{a(a+2b-c)}{b^2+c^2}-\frac{b(b+2a-c)}{a^2+c^2}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(a^3+b^3-c^3+3a^2b+3ab^2-a^2c-b^2c-abc+ac^2+bc^2)}{(a^2+c^2)(b^2+c^2)}\geq0$$ and the rest is very ugly.

I think the following way is better.

Let $a\geq b\geq c$.

Hence, $$\sum_{cyc}\frac{a^3+abc}{b^2+c^2}-a-b-c=\sum_{cyc}\left(\frac{a^3+abc}{b^2+c^2}-\frac{b+c}{2}\right)=$$ $$=\frac{1}{2}\sum_{cyc}\frac{2a^3+2abc-b^3-c^2-b^2c-bc^2}{b^2+c^2}=$$ $$=\frac{1}{2}\sum_{cyc}\frac{a^3-b^3-(c^3-a^3)+bc(a-b-(c-a))}{b^2+c^2}=$$ $$=\frac{1}{2}\sum_{cyc}\frac{(a-b)(a^2+b^2+ab+bc)-(c-a)(a^2+c^2+ac+bc)}{b^2+c^2}=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a^2+b^2+ab+bc}{b^2+c^2}-\frac{a^2+b^2+ab+ac}{a^2+c^2}\right)=$$ $$=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(a^3+b^3+2a^2b+2ab^2+abc-c^3)}{(a^2+c^2)(b^2+c^2)}=$$ $$=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(a^3+b^3-c^3+3abc+2a^2b+2ab^2-2abc)}{(a^2+c^2)(b^2+c^2)}=$$ $$=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2((a+b-c)(a^2+b^2+c^2-ab+ac+bc)+2ab(a+b-c))}{(a^2+c^2)(b^2+c^2)}=$$ $$=\frac{\sum\limits_{cyc}(a^2+ab)}{2\prod\limits_{cyc}(a^2+b^2)}\cdot\sum_{cyc}(a-b)^2(a+b-c)(a^2+b^2)\geq$$ $$\geq\frac{\sum\limits_{cyc}(a^2+ab)}{2\prod\limits_{cyc}(a^2+b^2)}\cdot\left((a-c)^2(a+c-b)(a^2+c^2)+(b-c)^2(b+c-a)(b^2+c^2)\right)\geq$$ $$\geq\frac{\sum\limits_{cyc}(a^2+ab)}{2\prod\limits_{cyc}(a^2+b^2)}\cdot\left((b-c)^2(a-b)(a^2+c^2)+(b-c)^2(b-a)(b^2+c^2)\right)=$$ $$=\frac{(b-c)^2(a-b)^2(a+b)\sum\limits_{cyc}(a^2+ab)}{2\prod\limits_{cyc}(a^2+b^2)}\geq0.$$

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