How prove this inequality with $\sum\sqrt{a+b}\sum\frac{1}{\sqrt{2a+b+c}}\le\frac{9}{\sqrt{2}}$?

Question

I found the following inequality which I can't solve it.

Let $a,b,c>0$,prove or disprove $$(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})\left(\dfrac{1}{\sqrt{2a+b+c}}+\dfrac{1}{\sqrt{2b+c+a}}+\dfrac{1}{\sqrt{2c+a+b}}\right)\le \dfrac{9}{\sqrt{2}}\tag{1}$$

Special case： if we let $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}=1,a+b=x^2.b+c=y^2,c+a=z^2$,then inequality become flase inequality

in fact $(1)$ is different links inequality, so How to prove $(1)$? Thanks :)

Answer 1

Yes, it's true.

By C-S $$\left(\sum\limits_{cyc}\sqrt{a+b}\right)^2\leq\sum\limits_{cyc}(2a+2b+c)\sum_{cyc}\frac{a+b}{2a+2b+c}=5(a+b+c)\sum_{cyc}\frac{a+b}{2a+2b+c}$$ and $$\left(\sum\limits_{cyc}\frac{1}{\sqrt{2a+b+c}}\right)^2\leq\sum\limits_{cyc}\frac{3a+2b+2c}{2a+b+c}\sum\limits_{cyc}\frac{1}{3a+2b+2c}$$ Thus, it remains to prove that $$(a+b+c)\sum_{cyc}\frac{a+b}{2a+2b+c}\sum\limits_{cyc}\frac{3a+2b+2c}{2a+b+c}\sum\limits_{cyc}\frac{1}{3a+2b+2c}\leq\frac{81}{10}$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $$3u\sum\limits_{cyc}\frac{3u-a}{6u-a}\sum\limits_{cyc}\frac{6u+a}{3u+a}\sum\limits_{cyc}\frac{1}{6u+a}\leq\frac{81}{10}$$ or $$\frac{u(36u^3+15uv^2-w^3)(99u^3+12uv^2+w^3)(48u^2+v^2)}{(108u^3+18uv^2-w^3)(54u^3+9uv^2+w^3)(324u^3+18uv^2+w^3)}\leq\frac{1}{10}$$ or $f(w^3)\geq0$, where $$f(w^3)=(108u^3+18uv^2-w^3)(54u^3+9uv^2+w^3)(324u^3+18uv^2+w^3)-$$ $$-10u(36u^3+15uv^2-w^3)(99u^3+12uv^2+w^3)(48u^2+v^2)$$

But $$f'(w^3)=-(54u^3+9uv^2+w^3)(324u^3+18uv^2+w^3)+$$ $$+(108u^3+18uv^2-w^3)(324u^3+18uv^2+w^3)+(108u^3+18uv^2-w^3)(54u^3+9uv^2+w^3)+$$ $$+10u(99u^3+12uv^2+w^3)(48u^2+v^2)-10u(36u^3+15uv^2-w^3)(48u^2+v^2)=$$ $$=(324u^3+18uv^2+w^3)(54u^3+9uv^2-2w^3)+(108u^3+18uv^2-w^3)(54u^3+9uv^2+w^3)+$$ $$+10u(48u^2+v^2)(63u^3-3uv^2+2w^3)>0$$ which says that $f$ is an increasing function.

Id est, $f$ gets a minimal value for a minimal value of $w^3$,

which happens for equality case of two variables or for $w^3\rightarrow0^+$.

1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$ and $b=1$.

We need to prove that $$368a^6+844a^5-271a^4-1432a^3-271a^2+844a+368\geq0$$

Let $a^2+1=ka$. Hence, $k\geq 2$ and we need to prove that $$368(k^3-3k)+844(k^2-2)-271k-1432\geq0$$ or $$368k^3+844k^2-833k-3120\geq0$$ and since $g(k)=368k^3+844k^2-833k-3120$ is an increasing function on $[2,+\infty)$, it's enough to prove that $g(2)\geq0$ and indeed, $g(2)=1534>0$;

2. Let $b=c=1$.

We get $(a-1)^2(92a^4+514a^3+945a^2+628a+80)\geq0$.

Done!