Prove the inequality $\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32$ where $a,b,c$ are positive reals.

Question

Let $a,b,c$ be positive real numbers, prove that $$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32.$$

The problem is from an inequality handout. Here is my attempt to solve the problem:

I first rewrote the inequality as $$\sum_{cyc}\frac{a^2}{(b+c)\sqrt{b^2-bc+c^2}}\geq\frac 32.$$ Here I noticed that the inequality has some similarities with Nesbitt's inequality, because that says $$\sum_{cyc}\frac{a}{b+c}\geq\frac 32.$$ But I don't know how to use this to reach the given inequality.

I also used Cauchy-Schwarz and got $$\sum_{cyc}\frac{a^2}{(b+c)\sqrt{b^2-bc+c^2}}\geq\frac{(a+b+c)^2}{\sum_{cyc}(b+c)\sqrt{b^2-bc+c^2}}.$$ But this also makes the problem too complicated that involves too many square roots. I think the square root on the denominator should be removed first to prove the inequality. But I am unable to do that.

So, I need a solution to the problem. Also it would be helpful for me if the answerer provides some motivation for the solution.

Answer 1

Another way.

By Holder $$\left(\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\right)^2\sum_{cyc}a^2(b+c)(b^3+c^3)\geq(a^2+b^2+c^2)^3$$ and it's enough to prove that: $$4(a^2+b^2+c^2)^3\geq9\sum_{cyc}(a^2b+a^2c)(b^3+c^3)$$ or $$\sum_{cyc}(4a^6+3a^4b^2+3a^4c^2-9a^3b^2c-9a^3c^2b+8a^2b^2c^2)\geq0,$$ which is true by Schur, AM-GM, Schur and Muirhead: $$\sum_{cyc}(4a^6+3a^4b^2+3a^4c^2-9a^3b^2c-9a^3c^2b+8a^2b^2c^2)\geq$$ $$\geq\sum_{cyc}(7a^4b^2+7a^4c^2-9a^3b^2c-9a^3c^2b+4a^2b^2c^2)\geq$$ $$\geq\sum_{cyc}(14a^4bc-9a^3b^2c-9a^3c^2b+4a^2b^2c^2)=$$ $$=9abc\sum_{cyc}(a^3-a^2b-a^2c+abc)+5abc\sum_{cyc}(a^3-abc)\geq0.$$

Answer 2

Another way:

Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}-\frac{3}{2}=\sum_{cyc}\frac{4a^2}{2\sqrt{(b+c)^2(4b^2-4bc+4c^2)}}-\frac{3}{2}\geq$$ $$\geq\sum_{cyc}\left(\frac{4a^2}{(b+c)^2+4b^2-4bc+4c^2}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{8a^2-5b^2-5c^2+2bc}{5b^2-2bc+5c^2}=$$ $$=\frac{1}{2}\sum_{cyc}\frac{(a-b)(4a+5b-c)-(c-a)(4a+5c-b)}{5b^2-2bc+5c^2}=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{4a+5b-c}{5b^2-2bc+5c^2}-\frac{4b+5a-c}{5a^2-2ac+5c^2}\right)=$$ $$=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(20a^2+45ab+20b^2-13ac-13bc-3c^2)}{(5a^2-2ac+5c^2)(5b^2-2bc+5c^2)}\geq$$ $$\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(16(a+b)^2-13ac-13bc-3c^2)}{(5a^2-2ac+5c^2)(5b^2-2bc+5c^2)}=$$ $$=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(a+b-c)(16a+16b+3c)}{(5a^2-2ac+5c^2)(5b^2-2bc+5c^2)}\geq$$ $$\geq\frac{1}{2}\left(\frac{(a-c)^2(a+c-b)(16a+16c+3b)}{(5a^2-2ab+5b^2)(5b^2-2bc+5c^2)}+\frac{(b-c)^2(b+c-a)(16b+16c+3a)}{(5a^2-2ab+5b^2)(5a^2-2ac+5c^2)}\right)\geq$$ $$\geq\frac{1}{2}\left(\frac{(b-c)^2(a-b)(16b+16c+3a)}{(5a^2-2ab+5b^2)(5a^2-2ac+5c^2)}-\frac{(b-c)^2(a-b)(16b+16c+3a)}{(5a^2-2ab+5b^2)(5a^2-2ac+5c^2)}\right)=0.$$

Answer 3

By AM-GM, C-S, Schur and Muirhead we obtain: $$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}=\sum_{cyc}\frac{4a^2}{2\sqrt{(b+c)^2(4b^2-4bc+4c^2)}}\geq\sum_{cyc}\frac{4a^2}{(b+c)^2+4b^2-4bc+4c^2}=$$ $$=\sum_{cyc}\frac{4a^2}{5b^2-2bc+5c^2}=\sum_{cyc}\frac{4a^4}{5a^2b^2-2a^2bc+5a^2c^2}\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(10a^2b^2-2a^2bc)}=$$ $$=\frac{\sum\limits_{cyc}(4a^4+8a^2b^2)}{\sum\limits_{cyc}(10a^2b^2-2a^2bc)}\geq\frac{\sum\limits_{cyc}(a^4+3a^3b+3a^3c+8a^2b^2-3a^2bc)}{\sum\limits_{cyc}(10a^2b^2-2a^2bc)}\geq\frac{3}{2}.$$