Balkan Mathematical Olympiad Shortlisted Problems

The desired inequality holds if (and only if) the following one does $$2{\left(\frac{a^2+b^2}{a+b}\right)}^3\ge a^3+b^3(\ast {)}_{a,b}$$ for all $a,b>0$. Indeed, $(\ast {)}_{a,b}$ is implied by the problem statement by setting $b=c$. Conversely, we recover the problem statement by summing $(\ast {)}_{a,b}$, $(\ast {)}_{b,c}$ and $(\ast {)}_{c,a}$. We now prove $(\ast {)}_{a,b}$ by observing that: $$2(a^2+b^2{)}^3-(a^3+b^3)(a+b{)}^3=(a-b{)}^4(a^2+ab+b^2)\ge 0.$$

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Alternative solution.

One sees that the expressions in the statement are related by the following identity: $$3(a^2+b^2)(a+b)=2(a^3+b^3)+(a+b{)}^3$$ Combining this with the AM-GM inequality below $$4{\left(\frac{a^2+b^2}{a+b}\right)}^3+\frac{(a+b{)}^3}{2}+\frac{(a+b{)}^3}{2}\ge 3\cdot \frac{a^2+b^2}{a+b}\cdot (a+b{)}^2=3(a^2+b^2)(a+b)$$ we recover the inequality $(\ast {)}_{a,b}$ as in the first solution.

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Alternative solution.

Let $(a,b,c)=(1,1,x)$ for $x>0$. The inequality then becomes $$2{\left(\frac{1+x^2}{1+x}\right)}^3\ge 1+x^3.$$ The key insight is that this inequality being true for all $x>0$ is equivalent to the original inequality being true for all $a,b,c>0$. Indeed, plugging in $x=a/b$, $x=b/c$, and $x=c/a$ respectively, we get $$2{\left(\frac{a^2+b^2}{a+b}\right)}^3\ge a^3+b^3$$ $$2{\left(\frac{b^2+c^2}{b+c}\right)}^3\ge b^3+c^3$$ $$2{\left(\frac{c^2+a^2}{c+a}\right)}^3\ge c^3+a^3$$ Summing these three inequalities yields the desired original one. Now, we finish by $$2(1+x^2{)}^3-(1+x^3)(1+x{)}^3=(x-1{)}^4(x^2+x+1)\ge 0.$$