Show that $\sqrt{4x^2+12xy+5y^2}+\sqrt{4y^2+12yz+5z^2}+\sqrt{4z^2+12xz+5x^2}<=13\sqrt{21}$

Question

question

Let the real numbers $x,y,z$ be positive nonzeros that fulfil the condition $x+y+z=13$. Show that $\sqrt{4x^2+12xy+5y^2}+\sqrt{4y^2+12yz+5z^2}+\sqrt{4z^2+12xz+5x^2}<=13\sqrt{21}$.

idea

This was my idea but it didn't lead to a good answer. Still, the number is near and I think we can use what I thought of.

Hope one of you can help me!

Answer 1

Another way.

By C-S $$\sum_{cyc}\sqrt{4x^2+12xy+5y^2}\leq\sqrt{\sum_{cyc}(2x+y)\sum_{cyc}(2x+5y)}=\sqrt{21}(x+y+z)=13\sqrt{21}.$$

C-S inequality it's the following:

For any reals $a_i$ and $b_i$ prove that: $$\left(a_1^2+a_2^2+...+a_n^2\right)(b_1^2+b_2^2+...+b_n^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2.$$ In our case $n=3$: $$\sum_{cyc}(2x+y)\sum_{cyc}(2x+5y)=\sum_{cyc}\left(\sqrt{2x+y}\right)^2\sum_{cyc}\left(\sqrt{2x+5y}\right)^2\geq$$ $$\geq\left(\sum_{cyc}\left(\sqrt{2x+y}\cdot\sqrt{2x+5y}\right)\right)^2=\left(\sum_{cyc}\sqrt{(2x+y)(2x+5y}\right)^2,$$ which says $$\sum_{cyc}\sqrt{4x^2+12xy+5y^2}\leq\sqrt{\sum_{cyc}(2x+y)\sum_{cyc}(2x+5y)}.$$

Answer 2

We need to prove that: $$\sum_{cyc}\sqrt{4x^2+12xy+5y^2}\leq\sqrt{21}(x+y+z)$$ or $$\sum_{cyc}\left(\sqrt{21}x-\sqrt{4x^2+12xy+5y^2}\right)\geq0$$ or $$\sum_{cyc}\frac{21x^2-(4x^2+12xy+5y^2)}{\sqrt{21}x+\sqrt{4x^2+12xy+5y^2}}\geq0$$ or $$\sum_{cyc}\frac{(x-y)(17x+5y)}{\sqrt{21}x+\sqrt{4x^2+12xy+5y^2}}\geq0$$ or $$\sum_{cyc}\left(\frac{(x-y)(17x+5y)}{\sqrt{21}x+\sqrt{4x^2+12xy+5y^2}}-\frac{11}{\sqrt{21}}(x-y)\right)\geq0$$ or $$\sum_{cyc}\frac{(x-y)(21(6x+5y)^2-121(4x^2+12xy+5y^2)}{\left(\sqrt{21}x+\sqrt{4x^2+12xy+5y^2}\right)\left(\sqrt{21}(6x+5y)+11\sqrt{4x^2+12xy+5y^2}\right)}\geq0$$ or $$\sum_{cyc}\frac{(x-y)^2(17x+5y)}{\left(\sqrt{21}x+\sqrt{4x^2+12xy+5y^2}\right)\left(\sqrt{21}(6x+5y)+11\sqrt{4x^2+12xy+5y^2}\right)}\geq0$$ and we are done!

Answer 3

Another way.

Since $f(x)=\sqrt{x}$ is a concave function, by Jensen we obtain: $$\sum_{cyc}\sqrt{4x^2+12xy+5y^2}=\sum_{cyc}\sqrt{(2x+y)(2x+5y)}=$$ $$=\sum_{cyc}\left(\frac{2x+y}{3(x+y+z)}\cdot\sqrt{\frac{9(x+y+z)^2(2x+y)(2x+5y)}{(2x+y)^2}}\right)\leq$$ $$\leq\sqrt{\sum_{cyc}\left(\frac{2x+y}{3(x+y+z)}\cdot\frac{9(x+y+z)^2(2x+y)(2x+5y)}{(2x+y)^2}\right)}=\sqrt{21}(x+y+z)=13\sqrt{21}.$$