Note that $\frac{a}{a^3+8}\leq\frac{6(a-1)}{9^2}+\frac{1}{9}$ because this is equivalent to $(a-1)^2(2a^2+5a+8)\geq 0$. Summing up the similar inequalities for $b,c,d$ we get the required result.

By AM-GM, $a^3+8\geq 9a^{\frac 13}$, thus the convexity of $x^{\frac 23}\ (x\geq 0)$, we have $L.H.S.\leq \frac 19(a^{\frac 23}+b^{\frac 23}+c^{\frac 23}+d^{\frac 23})$ $\leq \frac 49\left(\frac{a+b+c+d}{4}\right)^{\frac 23}=\frac 49.\ Q.E.D.$

Another way inspired from "kunny" from AM-GM $a^3+2\geq{3a}$ $\implies{a^3+8\geq{3a+6}}$ $\frac{a}{a^{3}+8} \leq{\frac{a}{3(a+2)}}$ $\frac{a}{a^{3}+8}+\frac{b}{b^{3}+8}+\frac{c}{c^{3}+8}+\frac{d}{d^{3}+8}\leq{\frac{1}{3}(\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}+\frac{d}{d+2})}$ $=\frac{4}{3}-\frac{2}{3}(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}+\frac{1}{d+2}\leq{\frac{4}{3}-\frac{2}{3}\cdot \frac{16}{12}}=\frac{4}{9}$ from C-S

Nice Solution, mcrasher. The solution is getting easier like $Differentiation\rightarrow Jensen\rightarrow Cauchy-Schwarz$, What's the next? Maybe, $AM-GM-HM?$

$\sum_{cyc}\frac{a}{a^3 +8} \leqslant \frac{4}{9} \iff \sum_{cyc}\frac{(a-1)(8-a-a^2)}{9(a^3 +8)} \leqslant 0$. Observe that $f(x)=x-1$ is increasing whilst $f(x)=\frac{8-x-x^2}{9(x^3 +8)}$ is decreasing on $[0,1]$, hence by Chebyshev's inequality $ \sum_{cyc}\frac{(a-1)(8-a-a^2)}{9(a^3 +8)} \leqslant \frac{1}{4}\left(a+b+c+d-4\right)\left(\sum_{cyc}\frac{8-a-a^2}{9(a^3 +8)}\right)=0$, as desired.

Observe that $ f(x)=x-1 $ is$ f(x)=\frac{8-x-x^{2}}{9(x^{3}+8)} $ increasing whilst is decreasing on $ [0,4] $, hence by Chebyshev's inequality $ \sum_{cyc}\frac{(a-1)(8-a-a^{2})}{9(a^{3}+8)}\leqslant\frac{1}{4}\left(a+b+c+d-4\right)\left(\sum_{cyc}\frac{8-a-a^{2}}{9(a^{3}+8)}\right)=0 $, as desired.

Tangent Line method Let $f(x)=\frac{x}{x^3+8}$ $f(x)\leq \frac{2}{27}x+\frac{1}{27}$ $\Leftrightarrow$ $(x-1)^2(2x^2+5x+8)\geq 0$ This holds when $0\leq x\leq 4$.So $\sum_{cyc}f(a)\leq \frac{2}{27}\cdot 4+\frac{4}{27}=\frac{4}{9}$

nice solution,Takeya.O

Let $a,b,c,d$ be non-negative reals such that $a+b+c+d=4$. Prove that$$\frac{a^3}{a^3+1}+\frac{b^3}{b^3+1}+\frac{c^3}{c^3+1}+\frac{d^3}{d^3+1}\leq\frac{16}5 $$bahdsn：

Attachments:

Note that $a^3 + 8 \ge 9\sqrt[9]{a^3}$ so $\frac{a}{a^3+8}+\frac{b}{b^3+8}+\frac{c}{c^3+8}+\frac{d}{d^3+8} \le \frac{1}{9}(\sqrt[3]{a^2} + \sqrt[3]{b^2} + \sqrt[3]{c^2} + \sqrt[3]{d^2})$ so we need to prove $\sqrt[3]{a^2} + \sqrt[3]{b^2} + \sqrt[3]{c^2} + \sqrt[3]{d^2} \le 4$. Note that $64 = (a+b+c+d)(a+b+c+d)(1+1+1+1) \ge (\sqrt[3]{a^2} + \sqrt[3]{b^2} + \sqrt[3]{c^2} + \sqrt[3]{d^2})^3$. we're Done.

Let $a,b,c,d$ be positive reals such that $a+b+c+d=4$. Prove that $$\frac{a}{a^3+5}+\frac{b}{b^3+5}+\frac{c}{c^3+5}+\frac{d}{d^3+5}\le\frac{2}{3}$$(2022 Croatia)

S

$$\frac{ab}{a^3+5}+\frac{bc}{b^3+5}+\frac{cd}{c^3+5}+\frac{da}{d^3+5}\le\frac{2}{3}$$ h

WakeUp wrote: Let $a,b,c,d$ be non-negative reals such that $a+b+c+d=4$. Prove the inequality \[\frac{a}{a^3+8}+\frac{b}{b^3+8}+\frac{c}{c^3+8}+\frac{d}{d^3+8}\le\frac{4}{9}\] Who is the author?