# Prove that: $\sum\limits_{cyc}\frac{1}{\sqrt{2a^2+5ab+2b^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}$

Let $$a$$, $$b$$ and $$c$$ be non-negative numbers such that $$ab+ac+bc\neq0$$. Prove that: $$\dfrac{1}{\sqrt{2a^2+5ab+2b^2}}+\dfrac{1}{\sqrt{2b^2+5bc+2c^2}}+\dfrac{1}{\sqrt{2c^2+5ca+2a^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}.$$

I solved this problem by Holder: $$\left(\sum_{cyc}\dfrac{1}{\sqrt{2a^2+5ab+2b^2}}\right)^2\sum_{cyc}\frac{(a+b)^3}{(2a^2+5ab+2b^2)^2}\geq\left(\sum_{cyc}\frac{a+b}{2a^2+5ab+2b^2}\right)^3$$ and it remains to prove that $$(ab+ac+bc)\left(\sum_{cyc}\frac{a+b}{2a^2+5ab+2b^2}\right)^3\geq3\sum_{cyc}\frac{(a+b)^3}{(2a^2+5ab+2b^2)^2},$$ which is true by BW and by using computer.

In this topic https://artofproblemsolving.com/community/c6h542992 there is a proof (from gxggs), but it's very very complicated.

I found another way, a smooth enough, but it's still a very hard solution.

I am looking for a nice proof by hand, for which there is a possibility to find this proof during a competition.

Thank you!

• Happy to see you back, Michael. Sep 12 at 11:58
• Side note: Better to just stick to positive numbers? EG You need $ab\neq0$. Sep 12 at 14:45

Remark: I give a proof using the so-called isolated fudging.

It suffices to prove that $$\frac{\sqrt{\frac{ab + bc + ca}{3}}}{\sqrt{2a^2 + 5ab + 2b^2}}\ge \frac{8c^2 + 9(a + b)c + 8ab}{8(a^2 + b^2 + c^2) + 26(ab + bc + ca)}. \tag{1}$$ Note: Taking cyclic sum on (1), we get the desired inequality.

If $$c = 0$$, it is easy.

If $$c > 0$$, WLOG, assume that $$c = 1$$. Let $$p = a + b, ~ q = ab$$. Then $$0 \le q \le p^2/4$$. It suffices to prove that $$\frac{\sqrt{(q + p)/3}}{\sqrt{2(p^2 - 2q) + 5q}} \ge \frac{8 + 9p + 8q}{8(p^2 - 2q + 1) + 26(q + p)}.$$ Squaring both sides, it suffices to prove that \begin{align*} &-92\,{q}^{3}+ \left( -224\,{p}^{2}+188\,p-224 \right) {q}^{2}+ \left( 64\,{p}^{4}-288\,{p}^{3}+313\,{p}^{2}+144\,p-128 \right) q \\ &\quad +64\,{p}^{5} -70\,{p}^{4}-60\,{p}^{3}+32\,{p}^{2}+64\,p \ge 0. \end{align*} Denote LHS by $$f(q)$$. We have $$f''(q) = - 448p^2 + 376p - 448 - 552q$$. So, we have $$f''(q) < 0$$ on $$q \ge 0$$. So, $$f(q)$$ is concave on $$q \ge 0$$. Also, we have $$f(0) = 64p^5 - 70p^4 - 60p^3 + 32p^2 + 64p \ge 0$$ and $$f(p^2/4) = \frac{1}{16}\,p \left( 9\,{p}^{3}+96\,{p}^{2}+256\,p+256 \right) \left( p-2 \right) ^{2} \ge 0$$. Thus, $$f(q) \ge 0$$ for all $$q\in [0, p^2/4]$$.

We are done.

• Thank you! I have found a very easy proof by $uvw$. Sep 13 at 2:03
• @MichaelRozenberg Nice. uvw is very powerful. Sep 13 at 2:09
• I have seen this technique a few times now and wondering if you have a general technique to find the lower bounds that sum up to $1.$ Sep 13 at 2:33
• @dezdichado If quadratic fraction actually work i.e. $f(a,b,c)\ge\frac{p_1a^2+p_2b^2+p_3c^2+p_4ab+p_5bc+p_6ca}{a^2+b^2+c^2+q(ab+bc+ca)}$, we need to find the coefficients which satisfies such as $f(a_0, b_0, c_0)$ ($a_0, b_0, c_0$ is the equality case for the inequality $f(a,b,c) + f(b,c,a) + f(c,a,b) \ge 1$, also the stationary condition $f'(a_0, b_0, c_0) = 0$, also sum to 1. Sep 13 at 3:09
• @dezdichado If $f(a, b, c)$ is symmetric about $a, b$, we can simplify something such as in this proof, $p_1 = p_2$, $p_5 = p_6$, etc. Sep 13 at 3:11

I found the following solution.

We need to prove that: $$\sum_{cyc}\left(\frac{1}{\sqrt{2a^2+5ab+2b^2}}-\frac{1}{\sqrt{3(ab+ac+bc)}}\right)\geq0$$ or $$\sum_{cyc}\frac{3ac+3bc-2a^2-2b^2-2ab}{2a^2+5ab+2b^2+\sqrt{3(ab+ac+bc)(2a^2+5ab+2b^2)}}\geq0$$ or $$\sum_{cyc}\frac{(c-a)(2a+b)-(b-c)(2b+a)}{2a^2+5ab+2b^2+\sqrt{3(ab+ac+bc)(2a^2+5ab+2b^2)}}\geq0$$ or $$\sum_{cyc}(a-b)\left(\tfrac{2b+c}{2b^2+5bc+2c^2+\sqrt{3(ab+ac+bc)(2b^2+5bc+2c^2)}}-\tfrac{2a+c}{2a^2+5ac+2c^2+\sqrt{3(ab+ac+bc)(2a^2+5ac+2c^2)}}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2\left((2a+c)(2b+c)-\tfrac{3c\sqrt{3(ab+ac+bc)(2a+c)(2b+c)}}{\sqrt{(a+2c)(2b+c)}+\sqrt{(2a+c)(b+2c)}}\right)\left(2a^2+5ab+2b^2+\sqrt{3(ab+ac+bc)(2a^2+5ab+2b^2)}\right)\geq0.$$ Now, let $$a\geq b\geq c$$.

Thus, since $$(2a+c)(2b+c)-\tfrac{3c\sqrt{3(ab+ac+bc)(2a+c)(2b+c)}}{\sqrt{(a+2c)(2b+c)}+\sqrt{(2a+c)(b+2c)}}\geq0,$$ $$(2a+b)(2c+b)-\tfrac{3b\sqrt{3(ab+ac+bc)(2a+b)(2c+b)}}{\sqrt{(a+2b)(2c+b)}+\sqrt{(2a+b)(c+2b)}}\geq0$$ and $$b(a-c)\geq a(b-c),$$ it remains to prove that: $$a^2\left((2a+b)(2c+b)-\tfrac{3b\sqrt{3(ab+ac+bc)(2a+b)(2c+b)}}{\sqrt{(a+2b)(2c+b)}+\sqrt{(2a+b)(c+2b)}}\right)\left(2a^2+5ac+2c^2+\sqrt{3(ab+ac+bc)(2a^2+5ac+2c^2)}\right)+$$ $$+b^2\left((2b+a)(2c+a)-\tfrac{3a\sqrt{3(ab+ac+bc)(2b+a)(2c+a)}}{\sqrt{(b+2a)(2c+a)}+\sqrt{(2a+b)(c+2a)}}\right)\left(2b^2+5bc+2c^2+\sqrt{3(ab+ac+bc)(2b^2+5bc+2c^2)}\right)\geq0.$$ But $$2a^2+5ac+2c^2+\sqrt{3(ab+ac+bc)(2a^2+5ac+2c^2)}\geq2b^2+5bc+2c^2+\sqrt{3(ab+ac+bc)(2b^2+5bc+2c^2)},$$ $$\sqrt{(a+2b)(2c+b)}+\sqrt{(2a+b)(c+2b)}=\sqrt{4b^2+5ab+4ac+5bc+2\sqrt{(2a^2+5ab+2b^2)(2c^2+5bc+2b^2)}}\geq$$ $$\geq\sqrt{4b^2+4ab+4ac+4bc+2\sqrt{(2a^2+4ab+2b^2)(2c^2+4bc+2b^2)}}=$$ $$=\sqrt{8(a+b)(b+c)}$$ and similarly $$\sqrt{(b+2a)(2c+a)}+\sqrt{(2a+b)(c+2a)}\geq\sqrt{8(a+b)(a+c)}.$$ Thus, it's enough to prove that: $$a^2(2a+b)(2c+b)+b^2(2b+a)(2c+a)\geq\tfrac{3ab\sqrt{3(ab+ac+bc)}}{\sqrt{8(a+b)}}\left(a\sqrt{\tfrac{2a+b)(2c+b)}{b+c}}+b\sqrt{\tfrac{(2b+a)(2c+a)}{a+c}}\right).$$ Now, by C-S $$a\sqrt{\tfrac{2a+b)(2c+b)}{b+c}}+b\sqrt{\tfrac{(2b+a)(2c+a)}{a+c}}\leq\sqrt{(a+b)\left(\tfrac{a(2a+b)(2c+b)}{b+c}+\tfrac{b(2b+a)(2c+a)}{a+c}\right)}\leq$$ $$\leq\sqrt{(a+b)\left(\tfrac{a(2a+b)\frac{3}{2}(c+b)}{b+c}+\tfrac{b(2b+a)\frac{3}{2}(c+a)}{a+c}\right)}=\sqrt{3(a+b)(a^2+ab+b^2)}.$$ Id est, it's enough to prove that: $$8\left(a^2(2a+b)(2c+b)+b^2(2b+a)(2c+a)\right)^2\geq81a^2b^2(ab+ac+bc)(a^2+ab+b^2),$$ which is true because $$8\left(a^2(2a+b)(2c+b)+b^2(2b+a)(2c+a)\right)^2=$$ $$=8\left(2a^2b^2+2a^3b+2ab^3+2a^2bc+2b^2ac+4a^3c+4b^3c\right)^2\geq$$ $$\geq8\left(2a^2b^2+2a^3b+2ab^3+2a^2bc+2b^2ac+4a^2bc+4b^2ac\right)^2=$$ $$=32a^2b^2(a^2+ab+b^2+3ac+3bc)^2\geq$$ $$\geq32a^2b^2((a^2+ab+b^2)^2+6(a^2+ab+b^2)(ac+bc))=$$ $$=32a^2b^2(a^2+ab+b^2)(a^2+ab+b^2+6(ac+bc))\geq$$ $$\geq32a^2b^2(a^2+ab+b^2)(3ab+3(ac+bc))=$$ $$=96a^2b^2(a^2+ab+b^2)(ab+ac+bc)\geq$$ $$\geq81a^2b^2(a^2+ab+b^2)(ab+ac+bc)$$ and we are done!

I think, the best way it's the following.

By the gxggs's idea (see here https://artofproblemsolving.com/community/c6h542992p3142759 ) it's enough to prove that $$\sum_{cyc}{\frac{12(a+b)}{17a^2+38ab+17b^2}}\geq \sqrt{\frac{3}{ab+bc+ca}}.$$ Now, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$ and $$abc=w^3$$.

Thus, since for $$(a,b,c)=(1,\zeta,\zeta^2)$$, where $$\zeta^2+\zeta+1=0$$, we have $$w^3\neq0$$, $$u=v^2=a^2+ab+b^2=a^2+ac+c^2=b^2+bc+c^2=0$$ and $$\prod_{cyc}(17a^2+38ab+17b^2)=21^3w^6,$$ we need to prove that $$f(w^3)\geq0,$$ where $$f$$ is a concave function,

which by $$uvw$$ (about $$uvw$$ see here: https://artofproblemsolving.com/community/c6h278791 ) says that it's enough to prove $$\sum_{cyc}{\frac{12(a+b)}{17a^2+38ab+17b^2}}\geq \sqrt{\frac{3}{ab+bc+ca}}$$ in the following cases.

1. Two variables are equal (it's enough to assume $$b=c=1$$);

2. $$w^3=0$$ (it's enough to assume $$c=0$$ and $$b=1$$),

which easy to check.

• You may separate two parts using <hr> before "I think, the best way it's the following." Or, move the uvw solution to another one. Sep 18 at 2:38