SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO

Let $a,b,c\in (0,\infty )$ and $a+b+c=3$.

By the Power Mean inequality, since $a,b,c>0$ and $a+b+c=3$: $$a^5+b^5+c^5\ge 3{\left(\frac{a+b+c}{3}\right)}^5=3\cdot 1^5=3.$$

Also, by the Cauchy-Schwarz inequality: $$ab+bc+ca\le \frac{(a+b+c{)}^2}{3}=\frac{9}{3}=3.$$ But we need a lower bound for $ab+bc+ca$.

Since $a,b,c>0$ and $a+b+c=3$, the minimum of $ab+bc+ca$ occurs when two variables are equal and the third is as small as possible (approaching $0$). Let $c\to 0$, $a+b\to 3$, $a,b>0$.

Then $ab+bc+ca\to ab$ as $c\to 0$. But $a+b=3$, so $ab$ is maximized when $a=b=1.5$, $ab=2.25$.

But for the minimum, $ab$ is minimized when $a\to 0$, $b\to 3$, $ab\to 0$. So $ab+bc+ca\to 0$ as one variable approaches $0$.

But let's check the minimum value of $a^5+b^5+c^5+8(ab+bc+ca)$ when $a,b,c>0$, $a+b+c=3$.

Let us try $a=b=c=1$: $$a^5+b^5+c^5+8(ab+bc+ca)=3\cdot 1^5+8(1\cdot 1+1\cdot 1+1\cdot 1)=3+8\cdot 3=3+24=27.$$

Now, try $a=0$, $b=0$, $c=3$ (but $a,b,c>0$): Let $a\to 0$, $b\to 0$, $c\to 3$: $$a^5+b^5+c^5+8(ab+bc+ca)\to 0+0+243+8(0+0+0)=243.$$ But $a,b,c>0$, so the minimum occurs when $a=b=c=1$.

Therefore, $$a^5+b^5+c^5+8(ab+bc+ca)\ge 27.$$ with equality when $a=b=c=1$.

Thus, the statement is proved.