How to prove $\frac{1}{\sqrt{2a^2+2bc+17}}+\frac{1}{\sqrt{2b^2+2ca+17}}+\frac{1}{\sqrt{2c^2+2ab+17}}\le \frac{\sqrt{21}}{7}$

Question

Let $a,b,c\ge 0: a+b+c=3.$ Prove that $$ \frac{3\sqrt{86}}{43}\le \frac{1}{\sqrt{2a^2+2bc+17}}+\frac{1}{\sqrt{2b^2+2ca+17}}+\frac{1}{\sqrt{2c^2+2ab+17}}\le \frac{\sqrt{21}}{7}$$ I tried to prove stronger inequaliy by C-S $$\frac{1}{2a^2+2bc+17}+\frac{1}{2b^2+2ca+17}+\frac{1}{2c^2+2ab+17}\le \frac{1}{7}$$But when I check $a=b=0$ the inequality leads to reverse one $$\frac{2}{17}+\frac{1}{18+17}=\frac{87}{595}>\frac{1}{7}$$ Also by Hölder $$\sum_{cyc}\frac{1}{\sqrt{2a^2+2bc+17}}=\sum_{cyc}\frac{\sqrt{a}}{\sqrt{2a^2+2bc+17}}.\frac{1}{\sqrt{a}}\le \sqrt{\sum_{cyc}\frac{a}{2a^2+2bc+17}.\sum_{cyc}\frac{1}{a}}\le \frac{\sqrt{21}}{7}$$ by squaring, we need to prove $$\sum_{cyc}\frac{a}{2a^2+2bc+17}.\sum_{cyc}\frac{1}{a}\le \frac{9}{21}$$ I'm stuck here. Hope to see some helps. Thank you for your help.

Update. Now I see that $$\sum_{cyc}\frac{a}{2a^2+2bc+17}.\sum_{cyc}\frac{1}{a}\le \frac{9}{21}$$is wrong when $a=b=0.$

Also, I check @TATAbox comment that$$\frac{1}{\sqrt{2a^2+2bc+17}}+\frac{1}{\sqrt{2b^2+2ca+17}}+\frac{1}{\sqrt{2c^2+2ab+17}}\ge \frac{3\sqrt{86}}{43}$$ is true

Answer 1

Let $$f(a,b,c)=\frac{1}{\sqrt{2 a^2+2 b c+17}}+\frac{1}{\sqrt{2 a c+2 b^2+17}}+\frac{1}{\sqrt{2 a b+2 c^2+17}}$$

With $c=3-a-b$ we get

$$g(a,b)=f(a,b,3-a-b)=\frac{1}{\sqrt{2 a^2+2 b (-a-b+3)+17}}+\frac{1}{\sqrt{2 a (-a-b+3)+2 b^2+17}}+\frac{1}{\sqrt{2 (-a-b+3)^2+2 a b+17}}$$

Now we want to find the maximum of $g(a,b)$:

$$\nabla _{\{a,b\}}g(a,b)=\begin{pmatrix} \frac{b-2 a}{\left(2 a^2-2 a b-2 (b-3) b+17\right)^{3/2}}+\frac{-2 a-3 b+6}{\left(2 a^2+6 a (b-2)+2 (b-6) b+35\right)^{3/2}}+\frac{2 a+b-3}{\left(-2 a b-2 (a-3) a+2 b^2+17\right)^{3/2}} \\ \frac{a+2 b-3}{\left(2 a^2-2 a b-2 (b-3) b+17\right)^{3/2}}+\frac{-3 a-2 b+6}{\left(2 a^2+6 a (b-2)+2 (b-6) b+35\right)^{3/2}}+\frac{a-2 b}{\left(-2 a b-2 (a-3) a+2 b^2+17\right)^{3/2}} \end{pmatrix}=0$$

The solution point here is $a= 1, b= 1$.
Now we have to check the Hessian Matrix at the solution point:

$$H |_{a=1,b=1}=\left( \begin{array}{cc} -\frac{4}{49 \sqrt{21}} & -\frac{2}{49 \sqrt{21}} \\ -\frac{2}{49 \sqrt{21}} & -\frac{4}{49 \sqrt{21}} \\ \end{array} \right)$$

is negative definite, hence we've got a maximum.

Finally we obtain the maximum value $$g(1,1)=f(1,1,1)=\sqrt{\frac{3}{7}}=\frac{\sqrt{21}}{7}$$

Other possible solution points are smaller

$$g(0,0)=g(3,0)=g(0,3)=\frac{1}{\sqrt{35}}+\frac{2}{\sqrt{17}}<\sqrt{\frac{3}{7}}$$

q.e.d.

Answer 2

Sketch of a proof for the right inequality.

We may use the so-called isolated fudging.

It suffices to prove that $$\frac{\sqrt{21}\,(a+b+c)/9}{\sqrt{2a^2+2bc+17(a+b+c)^2/9}} \le \frac{407a^2 + 693a(b + c) + 584(b^2 + c^2) + 504bc}{1575(a^2 + b^2 + c^2) + 1890(ab + bc + ca)}. \tag{1}$$ (Note: Take a cyclic sum on (1), we get the desired inequality.)

If $a = 0$, it is easy.

If $a > 0$, WLOG, assume that $a = 1$. Let $p = b + c, q = bc$. We have $p^2 \ge 4q$.

(1) is written as $$f(q) := \frac{407 + 693p + 584(p^2 - 2q) + 504q}{1575(1 + p^2 - 2q) + 1890(p + q)} - \frac{\sqrt{21}\,(1+p)/9}{\sqrt{2+2q+17(1+p)^2/9}} \ge 0.$$

It is easy to prove that $f(q)$ is concave. Also, we can prove that $f(0) \ge 0 $ and $f(p^2/4) \ge 0$. Thus, $f(q) \ge 0$ on $[0, p^2/4]$.

We are done.

Answer 3

Sketch of a proof.

We may use the two following facts below.

For the right side inequality.

Fact 1. For any $x,y,z\ge 0$ satisfying $$x^2+y^2+z^2+\frac{9}{10}x^2y^2z^2\le \frac{39}{10}, $$then $x+y+z\le 3.$

Applying this fact for $$x=\sqrt{\frac{21}{2a^2+2bc+17}};y=\sqrt{\frac{21}{2b^2+2ca+17}};z=\sqrt{\frac{21}{2c^2+2ab+17}}$$ we obtain $$x^2+y^2+z^2+\frac{9}{10}x^2y^2z^2\le \frac{39}{10} $$ and it gives $x+y+z\le 3.$

Equality holds iff $a=b=c=1.$

For the left side inequality.

Fact 2. For any $x,y,z\ge 0$ satisfying $$x^2y^2+y^2z^2+z^2x^2+3x^2y^2z^2\ge 6, $$then $x+y+z\ge 3.$

Applying this fact for $$x=\sqrt{\frac{43}{2(2a^2+2bc+17)}};y=\sqrt{\frac{43}{2(2b^2+2ca+17)}};z=\sqrt{\frac{43}{2(2c^2+2ab+17)}}$$ we obtain $$x^2y^2+y^2z^2+z^2x^2+3x^2y^2z^2\ge 6$$ and it gives $x+y+z\ge 3.$

Equality holds iff $a=b=\dfrac{3}{2}, c=0$ and its permutations.