TITLE: If $abc=1$ then $ \sum_{cyc} \frac{a^2\sqrt{3bc}}{\sqrt{2b^2-bc+2c^2}} \ge 3 $
QUESTION [5 upvotes]: (Đặng Trọng Nghĩa) Given three positive reals $a,b,c$ such that $abc=1$. Prove that:
$$ \sum_{cyc} \frac{a^2\sqrt{3bc}}{\sqrt{2b^2-bc+2c^2}} \ge 3$$
I tried AM-GM:
$$ \frac{a^2\sqrt{3bc}}{\sqrt{2b^2-bc+2c^2}} =\frac{3a}{\sqrt{3bc \left( 2b^2-bc+2c^2 \right)}} \ge \frac{3a}{\dfrac{3bc+2b^2-bc+2c^2}{2}}=\frac{3a}{b^2+bc+c^2}$$
Hence it remains to prove that:
$$ \sum_{cyc} \frac{a}{b^2+bc+c^2} \ge 1$$
Which is wrong for large enough $b,c$.
I then tried C-S:
$$ \frac{a^2\sqrt{3bc}}{\sqrt{2b^2-bc+2c^2}} = \frac{a^2}{\sqrt{\dfrac{2b^2-bc+2c^2}{3bc}}} = \frac{a^2}{\sqrt{\dfrac{a \left( 2b^2-bc+2c^2 \right)}{3}}} $$
So:
$$ \sum_{cyc} \frac{a^2\sqrt{3bc}}{\sqrt{2b^2-bc+2c^2}} \ge \frac{\left( a+b+c \right)^2}{\sum_{cyc} \sqrt{\dfrac{a \left( 2b^2-bc+2c^2 \right)}{3}}} $$
Using C-S again, we also have:
$$ \sum_{cyc} \sqrt{\dfrac{a \left( 2b^2-bc+2c^2 \right)}{3}} \le \sqrt{\left( \frac{a+b+c}{3} \right) \left( 4(a^2+b^2+c^2) - (ab+bc+ca) \right) } $$
Hence it remains to prove that:
$$ \frac{\sqrt{3(a+b+c)^3}}{\sqrt{4(a^2+b^2+c^2) - (ab+bc+ca)}} \ge 3$$
Which is wrong again at $b=c=2$.

REPLY [2 votes]: Holder in another way:
Using Holder, we have
$$\left(\sum_{\mathrm{cyc}} \frac{a^2\sqrt{3bc}}{\sqrt{2b^2 - bc + 2c^2}}\right)^2
\sum_{\mathrm{cyc}} 9a^3(2b^2 - bc + 2c^2)
\ge (3a^2 + 3b^2 + 3c^2)^3.$$
It suffices to prove that
$$(3a^2 + 3b^2 + 3c^2)^3
\ge 9 \sum_{\mathrm{cyc}} 9a^3(2b^2 - bc + 2c^2). \tag{1}$$
Hope to see a nice proof of (1). I have a complicated proof using the pqr method.

Let $p = a + b + c, q = ab + bc + ca, r = abc = 1$.
We have $p\ge 3$ and $p^2 \ge 3q$.
(1) is written as
$$27(p^2 - 2q)^3 \ge 81 p(2q^2 - 5pr)$$
or
$$27(p^2 - 2q)^3 \ge 81 p(2q^2 - 5p).$$
Using Schur's inequality $a^2(a - b)(a - c) + b^2(b - c)(b - a) + c^2(c - a)(c - b) \ge 0$ which is written as $p^4 - 5p^2q + 6pr + 4q^2 \ge 0$,
we have $p^4 - 5p^2q + 6p + 4q^2 \ge 0$ or
$\frac{1}{16}(5p^2 - 8q)^2\ge \frac{9p^4 - 96p}{16}$
which results in (using $p\ge 3$ and $p^2 \ge 3q$)
$$q \le \frac{5p^2 - \sqrt{9p^4 - 96p}}{8}.$$
Since $p^2 - 2\cdot \frac{5p^2 - \sqrt{9p^4 - 96p}}{8} = \frac{\sqrt{9p^4 - 96p} - p^2}{4} > 0$,
it suffices to prove that
$$27\left(p^2 - 2\cdot \frac{5p^2 - \sqrt{9p^4 - 96p}}{8}\right)^3 \ge 81 p\left(2\cdot \left(\frac{5p^2 - \sqrt{9p^4 - 96p}}{8}\right)^2 - 5p\right)$$
or
$$(81p^4 + 405p^3 - 648p)\sqrt{9p^4 - 96p} \ge 189p^6 + 1377p^5 - 1944p^3 - 10368p^2.$$
Since $81p^4 + 405p^3 - 648p > 0$, it suffices to prove that
$$(81p^4 + 405p^3 - 648p)^2(9p^4 - 96p) \ge (189p^6 + 1377p^5 - 1944p^3 - 10368p^2)^2$$
or
$$23328p^3(p - 3)(p^8 + 6p^7 - 36p^5 - 183p^4 + 432p^2 + 1728p + 576) \ge 0$$
which is true (using $p \ge 3$).
We are done.