# How to prove $\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac b{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}$ when $a,b,c>0$

I want to prove that for $$a,b,c>0$$ we have

$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}= \frac a{\sqrt{a^2+3b^2+3c^2}}+\frac{b}{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}.$$

My first attempt: By Cauchy-Schwarz we have $$\left(\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}\right)^2\le3\sum_{cyc}\frac{a^2}{a^2+3b^2+3c^2}$$ so we only need to prove that the right-hand side is always less than $$\frac{9}{7}$$, but this is false. Failed

Second attempt: By Cauchy-Schwarz

$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=\sum_{cyc} \frac 1{\sqrt{1+3\frac{b^2}{a^2}+3\frac{c^2}{a^2}}}\le\sum_{cyc} \frac{\sqrt 7}{1+3\frac{b}{a}+3\frac{c}a}$$

so it remains to prove that $$\sum_{cyc} \frac{a}{1+3b+3c}\le\frac37$$ but this is wrong for example for $$a=1,b=1,c=2$$. Failed

Third attempt: Let $$S=3(a^2+b^2+c^2)$$. We need to prove $$\sum_{cyc} \frac{a}{\sqrt{S-2a^2}}\le \frac37.$$ But $$x\mapsto \frac{x}{\sqrt{S -2x^2}}$$ is convex so Jensen has the wrong direction...

Just an observation, the equality is achieved not only for $$a=b=c$$, but also for $$(a^2 \colon b^2 \colon c^2) = (8\colon 1\colon 1)$$

$$\bf{Added:}$$

Define $$x=\frac{a^2}{a^2 + 3 b^2 + 3 c^2},\ y=\frac{b^2}{b^2 + 3 a^2 + 3 c^2},\ z=\frac{c^2}{c^2 + 3 a^2 + 3 b^2}$$

One checks (say by direct calculation) that $$28 x y z + 8(x y + x z + y z )+ x+y+z-1=0$$

So it is enough to show that the maximum of the function $$\sqrt{x}+\sqrt{y}+\sqrt{z}$$ on the part of the above surface in the first octant is $$\frac{3}{\sqrt{7}}$$. The Lagrange multiplier system $$28 x y z + 8(x y + x z + y z )+ x+y+z-1=0\\ t - x( 28 y z + 8 (y+z) + 1)^2 =0\\ t - y( 28 x z + 8 (x+z) + 1)^2 =0\\ t - z( 28 x y + 8 (x+y) + 1)^2 =0$$

is in fact not that hard to solve, if we use Groebner bases. First, by elimination we get the equation in $$t$$:

$$72313663744 t^7 - 207058475232 t^6 - 212349914280 t^5 + 806857109604 t^4 + 125825565483 t^3 - 784526490225 t^2=0$$ which factor nicely as $$t^2 (1372 t - 2025) (343 t - 729) (56 t + 81) (2744 t^2 - 1944 t - 6561)=0$$

Now one considers each of the possible positive values of $$t$$ and solves the system in $$x$$, $$y$$, $$z$$.

Case 1. $$343 t - 729=0$$. We get $$x=y=z=\frac{1}{7}$$

Case 2. $$1372 t - 2025=0$$. We get the solution $$x=\frac{4}{7}$$, $$y=z=\frac{1}{28}$$ and the cyclic permutation of it.

Case 3. $$t = \frac{81(6 + 19\sqrt{2})}{1372}$$

We get $$x=\frac{-2 + 3 \sqrt{2}}{7}$$, $$y=z = \frac{-2 + 3 \sqrt{2}}{28}$$ and the circular permutations.

Case 4. $$t=0$$ gives negative solutions, so we discard it

The inequality now follows, in the cases 1, 2, the functions takes the maximum value $$\frac{3}{\sqrt{7}}$$, in case 3 the value is smaller.

We need to prove that: $$\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\leq\frac{3}{\sqrt7},$$ where $$a$$, $$b$$ and $$c$$ are positive numbers.

Indeed, by C-S $$\left(\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\right)^2\leq\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\sum_{cyc}(17a+2b+2c).$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\leq\frac{3}{49(a+b+c)}.$$ Now, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$ and $$abc=w^3$$.

Thus, we need to prove that: $$\sum_{cyc}\frac{a}{(9u-2a)(6u+15a)}\leq\frac{1}{49u}$$ or $$49u\sum_{cyc}a(9u-2b)(9u-2c)(2u+5b)(2u+5c)\leq3\prod_{cyc}((9u-2a)(2u+5a)).$$ We'll prove that the last inequality is true even for any reals $$a$$, $$b$$ and $$c$$.

Indeed, since $$\sum\limits_{cyc}a(9u-2b)(9u-2c)(2u+5b)(2u+5c)$$ is a fifth degree polynomial,

the last inequality is equivalent to $$f(w^3)\geq0,$$ where $$f(w^3)=-3000w^6+A(u,v^2)w^3+B(u,v^2).$$ But $$f$$ is a concave function.

Thus, $$f$$ gets a minimal value for an extreme value of $$w^3$$, which happens for equality case of two variables.

Since the last inequality is an even degree, homogeneous and symmetrical, it's enough to assume $$b=c=1,$$ which gives $$\frac{a}{(a+6)(17a+4)}+\frac{2}{(3a+4)(2a+19)}\leq\frac{3}{49(a+2)}$$ or $$(a-1)^2(a-8)^2\geq0$$ and we are done!

• Can I ask: when you use Cauchy-Schwarz, do you look for $x,y,z$ such that: $$\frac{7a}{(a+3b+3c)(xa+yb+zc)^2}=\frac{7b}{(b+3a+3c)(xb+ya+zc)^2}=\frac{7c}{(c+3a+3b)(xc+ya+zb)^2}$$ for both equality cases? – LHF Feb 23 at 15:28
• @Atticus Yes, of course. Because our vectors should be parallel. – Michael Rozenberg Feb 23 at 15:33

Alternative proof:

Let $$x, y, z > 0$$ such that $$\frac{7a^2}{a^2 + 3b^2 + 3c^2} = x^2, \ \frac{7b^2}{b^2 + 3c^2 + 3a^2} = y^2, \ \frac{7c^2}{c^2 + 3a^2 + 3b^2} = z^2$$. It is not hard to obtain $$F(x, y, z) = 4 x^2 y^2 z^2+8 x^2 y^2+8 x^2 z^2+8 y^2 z^2+7 x^2+7 y^2+7 z^2-49 = 0.$$

It suffices to prove that if $$x, y, z > 0$$ and $$F(x,y,z) = 0$$, then $$x+y+z \le 3$$. It suffices to prove that if $$x, y, z > 0$$ and $$x + y + z > 3$$, then $$F(x, y, z) > 0$$. Note that $$F(\alpha x, \alpha y, \alpha z) > F(x, y, z)$$ for any $$\alpha > 1$$ and $$x, y, z > 0$$. Thus, it suffices to prove that if $$x, y, z > 0$$ and $$x+y+z = 3$$, then $$F(x, y, z) \ge 0$$.

We use pqr method. Let $$p = x + y + z = 3$$, $$q = xy + yz + zx$$ and $$r = xyz$$. From $$p^2 \ge 3q$$, we have $$q \le 3$$. Let $$q = 3(1-u^2)$$ for $$0 \le u\le 1$$. We have \begin{align} (x-y)^2(y-z)^2(z-x)^2 &= -4p^3r+p^2q^2+18pqr-4q^3-27r^2\\ & = 108u^6 - 27(3u^2+r-1)^2 \end{align} which results in $$108u^6 - 27(3u^2+r-1)^2 \ge 0$$ and hence $$r \le (2u+1)(1-u)^2 \le 3$$. Thus, we have \begin{align} F(x, y, z) &= 7p^2-16pr+8q^2+4r^2-14q-49 \\ &= 4(6-r)^2+72u^4-102u^2-100\\ &\ge 4(6 - (2u+1)(1-u)^2)^2 + 72u^4-102u^2-100\\ &= 2u^2(2u^2-4u+9)(2u-1)^2\\ &\ge 0. \end{align} We are done.

We need to prove that:

$$\sqrt{\frac{a}{a+3b+3c}}+\sqrt{\frac{b}{b+3c+3a}}+\sqrt{\frac{c}{c+3a+3b}}\leq\frac{3}{\sqrt7}$$

Let's normalize with $$a+b+c=3$$. Then the inequality is equivalent with:

$$\sqrt{\frac{a}{9-2a}}+\sqrt{\frac{b}{9-2b}}+\sqrt{\frac{c}{9-2c}}\leq \frac{3}{\sqrt{2}}$$

Without loss of generality suppose that $$a\le b\le c$$. Then we have $$a+b\leq 2$$ and we will prove:

$$\sqrt{\frac{a}{9-2a}}+\sqrt{\frac{b}{9-2b}} \leq \sqrt{\frac{2(a+b)}{9-a-b}}$$

Squaring twice, this is equivalent with:

$$\frac{(a-b)^2[729+81(a+b)^2-486(a+b)-16ab(a+b)]}{(9-2a)^2(9-2b)^2(9-a-b)^2}\geq 0$$

We have $$16ab(a+b)\leq 16(a+b)$$ and

$$729+81x^2-502x\geq 0, \text{ when }x \leq 2$$

It remains to prove that:

$$\sqrt{\frac{14(3-c)}{6+c}}+\sqrt{\frac{7c}{9-2c}}\leq 3$$

Squaring twice this is equivalent with:

$$\frac{81(c-1)^2 (12 - 5 c)^2}{(9 - 2 c)^2 (6 + c)^2}\geq 0$$

which gives the two equality cases.