$\frac{a^3}{a^2+1} \geq \frac{2a-1}{2}$ So $\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{2(a+b+c)-3}{2}=\frac{3}{2}$

Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=3$. Prove the inequality $$\frac{a^3}{a^2+b}+\frac{b^3}{b^2+c}+\frac{c^3}{c^2+a} \geq\frac{3}{2}\geq \frac{a^2}{a^3+1}+\frac{b^2}{b^3+1}+\frac{c^2}{c^3+1} $$$$\frac{a^3}{a^2+b+c}+\frac{b^3}{b^2+c+a}+\frac{c^3}{c^2+a+b} \geq 1 $$$$\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \leq\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} $$h

Just note that $\frac{a^3}{a^2+1} = a\cdot \left(1-\frac{1}{a^2+1}\right)$, and thus it boils down proving $\frac32 \ge \sum \frac{a}{a^2+1}$. But as $a/(a^2+1)\le 1/2$ we are done.

Let $a$, $b$ and $c$ be non-negative real numbers such that $a+b+c=3$. Prove that $$\frac{27}{10}\geq\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{3}{2}$$$$\frac{3}{2}\geq \frac{a^2}{a^3+1}+\frac{b^2}{b^3+1}+\frac{c^2}{c^3+1} \geq\frac{9}{28}$$

It is not difficult to prove that $$\frac{x^3}{x^2+1}\geq \frac{1}{2}+x-1$$ holds for all $x$, summing the result inequalities we are done.

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strong_boy wrote: Cute problem : We know $a^2 +1 \geq 2a$ . ( ) By ( ) and original problem we can see : $$\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{a^3}{2a}+\frac{a^3}{2a}+\frac{a^3}{2a} = a^2+b^2+c^2$$ Now we only need prove $a^2+b^2+c^2 \geq 3$ . and it is true beacuase $3(a^2+b^2+c^2) \geq (a+b+c)^2$ . $\blacksquare$ $X \ge Y$ doesn't mean $\frac{Z}{X} \ge \frac{Z}{Y}$, this is false.

Tangent line method... $\frac{x^3}{x^2+1}\geq x-\frac{1}{2}\Leftrightarrow \frac{1}{2}(x-1)^2\geq0$ $\sum_{cyc}\frac{a^3}{a^2+1}\geq\sum_{cyc}(a-\frac{1}{2})=\frac{3}{2}$

Write $\frac{a^3}{a^2+1}$ as $a-\frac{a}{a^2+1}$. Summing, we just want to prove $3-\sum_{cyc}\frac{a}{a^2+1}\geq \frac{3}{2}$ or rearranging, $\sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$. We claim $\frac{a}{a^2+1}\leq \frac{1}{2}$. This is trivial as upon rearranging, $(a-1)^2\geq 0$. Summing finishes.

Let $ a,b,c >0 $ and $ a+b+c=3$. Prove that $$\frac{a^3}{a^2+2}+\frac{ b^3}{b^2+8}+\frac{c^3}{c^2+2} \geq \frac{27}{41} $$$$\frac{a^3}{a^2+2}+\frac{ b^3}{b^2+3}+\frac{c^3}{c^2+2} \geq \frac{27(5-\sqrt 6)}{76} $$

sqing wrote: Let $a$, $b$ and $c$ be non-negative real numbers such that $a+b+c=3$. Prove that $$\frac{27}{10}\geq\frac{a^3}{a^2+1}+\frac{b^3}{b^2+1}+\frac{c^3}{c^2+1} \geq \frac{3}{2}$$$$\frac{3}{2}\geq \frac{a^2}{a^3+1}+\frac{b^2}{b^3+1}+\frac{c^2}{c^3+1} \geq\frac{9}{28}$$ The left hand side of the second inequality is due to tangent line trick. RTP $f\left(x\right)-f\left(1\right)-\left(x-1\right)f'\left(1\right)\leq 0$ if $f\left(x\right)=\frac{x^{2}}{x^{3}+1}$. Rearrange as showing $x^{4}+x^{3}-4x^{2}+x+1\geq 0$. Standard calc exercise. Derivative tells us local minimum at $x=1$, other minimums are at negative $x$ values. Easy to check that this gives $1+1-4+1+1=0$ as the $y$ coordinate hence everything else is just nonnegative.