<s>[INST] symetric inequality for a rational function of three variables $a,b,c$ are positive real numbers such that $a+b+c=1$. Prove that: $$\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2} \geqslant \dfrac{1}{2} $$ I have tried with Cauchy-Schwarz inequality in Engel form...
 [/INST] $$\sum_{cyc}\frac{a^3}{a^2+b^2}-\frac{1}{2}=\sum_{cyc}\frac{a^3}{a^2+b^2}-\frac{a+b+c}{2}=\sum_{cyc}\left(\frac{a^3}{a^2+b^2}-\frac{a}{2}\right)=$$
$$=\sum_{cyc}\frac{a^3-ab^2}{2(a^2+b^2)}=\sum_{cyc}\left(\frac{a^3-ab^2}{2(a^2+b^2)}-\frac{a-b}{2}\right)=\sum_{cyc}\frac{(a-b)^2b}{2(a^2+b^2)}\geq0.$$
Done!
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