MOP 2011 inequality

Question

If $a,b,c$ are positive integers prove that

$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$

My attempt:

I tried to split inequality and prove it bit by bit

$9(abc)^{1/3} \le 3(a+b+c)$ It suffices to prove that

$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} \le a+b+c$

After trying out few values I realized it is wrong.So I decided to tackle it all at your once. I tried few well known inequalities and played around but nothing came useful out of it. Any help??

Answer 1

Another way.

By AM-GM $$\sum_{cyc}\sqrt{a^2-ab+b^2}\leq\sum_{cyc}\left(\frac{a^2-ab+b^2}{a+b}+\frac{a+b}{4}\right)=$$ $$=\sum_{cyc}\left(\frac{a^2+2ab+b^2-3ab}{a+b}+\frac{a+b}{4}\right)=\frac{5}{2}(a+b+c)-3\sum_{cyc}\frac{ab}{a+b}=$$ $$=4(a+b+c)-3\sum_{cyc}\left(\frac{ab}{a+b}+\frac{a+b}{4}\right)\leq$$ $$\leq 4(a+b+c)-3\sum_{cyc}\sqrt{ab}\leq4(a+b+c)-9\sqrt[3]{abc}.$$

Answer 2

By Jensen $$\sum_{cyc}\sqrt{a^2-ab+b^2}\leq3\sqrt{\frac{\sum\limits_{cyc}(a^2-ab+b^2)}{3}}=\sqrt{3\sum_{cyc}(2a^2-ab)}.$$ Thus, it's enough to prove that: $$\sqrt{3\sum_{cyc}(2a^2-ab)}+9\sqrt[3]{abc}\leq4(a+b+c).$$ Now, by $uvw$ (see here: https://artofproblemsolving.com/community/c6h278791 )

it's enough to prove the last inequality for equality case of two variables.

Can you end it now?

Answer 3

Using $\sum $ to denote cyclic sums, \begin{align} \sum \sqrt{a^2-ab+b^2} \;+ 9\sqrt[3]{abc} &\leqslant \sum \left(\sqrt{a^2-ab+b^2}+3\sqrt{ab}\right) \tag{*}\\ &\leqslant \sum 2(a+b) \tag{**}\\ &=4\sum a \end{align}

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Notes: * AM-GM inequality ** Jensen's inequality with the concave function $t \mapsto\sqrt{t}$

Answer 4

As we can see that the term $9(abc)^{\frac{1}{3}}$ is difficult to deal with, we consider the following inequality $$-\sqrt{(a^2-ab+b^2)} -\sqrt{(b^2+c^2-bc)} -\sqrt{(a^2+c^2-ac)} +4(a+b+c)\ge 9(abc)^{1/3}.$$ Then we consider $$\frac{a+b}{2}-\sqrt{(a^2-ab+b^2)}=\frac{(\frac{a+b}{2})^2-(a^2-ab+b^2)}{\frac{a+b}{2}+\sqrt{(a^2-ab+b^2)}},$$the numerator is non-positive, and the denominator $$\frac{a+b}{2}+\sqrt{(a^2-ab+b^2)}\ge \frac{a+b}{2}+\frac{a+b}{2}.$$So we have $$\frac{(\frac{a+b}{2})^2-(a^2-ab+b^2)}{\frac{a+b}{2}+\sqrt{(a^2-ab+b^2)}}\ge \frac{(\frac{a+b}{2})^2-(a^2-ab+b^2)}{a+b}=\frac{3ab}{a+b}-\frac{3}{4}(a+b).$$ Hence $$-\sqrt{(a^2-ab+b^2)} -\sqrt{(b^2+c^2-bc)} -\sqrt{(a^2+c^2-ac)} +4(a+b+c)\ge \sum_{cyc} \frac{3ab}{a+b}+\frac{3}{2}\sum a.$$Then you just need to use AM-GM inequality to prove it is greater than $9(abc)^{1/3}$.

One of my schoolmate of lower grade also finds a solution(before me).With the permission of him, I post his solution as follows:$$2(a+b)-\sqrt{a^2+b^2-ab}=\frac{a+b}{2}+\frac{1}{2}\sqrt{a^2+b^2-ab}+\frac{3(a+b)}{2}-\frac{3}{2}\sqrt{a^2+b^2-ab}$$$$=\frac{a+b+\sqrt{a^2+b^2-ab}}{2}+\frac{3}{2}\frac{3ab}{a+b+\sqrt{a^2+b^2-ab}}\ge 3\sqrt{ab}$$ So sum up and use AM-GM, we can finish the proof.