Proving $\sqrt{a^{2}\left(b+c\right)+2}+\sqrt{b^{2}\left(c+a\right)+2}+\sqrt{c^{2}\left(a+b\right)+2}\ge \sqrt{6(ab+bc+ca+3)}$ for $a+b+c=3.$

Question

Let $a,b,c\ge 0: a+b+c=3$ then prove $$\sqrt{a^{2}\left(b+c\right)+2}+\sqrt{b^{2}\left(c+a\right)+2}+\sqrt{c^{2}\left(a+b\right)+2}\ge \sqrt{6(ab+bc+ca+3)}$$

I check $$\sqrt{a^{2}\left(b+c\right)+2}+\sqrt{b^{2}\left(c+a\right)+2}+\sqrt{c^{2}\left(a+b\right)+2}\ge 3\sqrt2\ge \sqrt{6(ab+bc+ca+3)}$$ which leads $ab+bc+ca<0:$ it is impossible.

Also, by AM-GM $ab+bc+ca\le \frac{(a+b+c)^2}{3}=3$ and we need to prove $$\sqrt{a^{2}\left(b+c\right)+2}+\sqrt{b^{2}\left(c+a\right)+2}+\sqrt{c^{2}\left(a+b\right)+2}\ge 6$$ Could you help me prove the stronger inequality? Thank you.

Answer 1

Some thoughts.

Firstly, we prove the following lemma.

Lemma. For any non-negative real numbers $x,y,z$ which is $$x^2y^2+y^2z^2+z^2x^2+3x^2y^2z^2\ge 6$$ then $$x+y+z\ge 3.$$ By applying the lemma with $$x=\dfrac{3\left[a^{2}\left(b+c\right)+2\right]}{2\left(ab+bc+ca+3\right)},\ y=\dfrac{3\left[b^{2}\left(c+a\right)+2\right]}{2\left(ab+bc+ca+3\right)},\ z=\dfrac{3\left[c^{2}\left(a+b\right)+2\right]}{2\left(ab+bc+ca+3\right)}$$ and we'll prove $$\frac{9}{4}\cdot\sum\frac{\left[a^2(b+c)+2\right]\left[b^2(c+a)+2\right]}{(ab+bc+ca+3)^2}+\frac{81}{8}\cdot \frac{\left[a^2(b+c)+2\right]\left[b^2(c+a)+2\right]\left[c^2(a+b)+2\right]}{(ab+bc+ca+3)^3}\ge 6.$$ Now, let $a+b+c=p=3,\ ab+bc+ca=q,\ abc=r.$ The inequality is rewritten as $$\displaystyle 6 q^{4} + 56 q^{3} - 72 q^{2} + 180 q - 27 r^{3} + r^{2} \left(99 q + 216\right) + r \left(- 36 q^{2} - 504 q - 540\right)\ge 0.$$ The last inequality is true.

Equality holds iff $(a,b,c)\sim(1,1,1)$ or $(a,b,c)\sim\left(\dfrac{3}{2},\dfrac{3}{2},0\right).$

Answer 2

I hope the following idea will help.

By Holder inequality$$\left(\sum_{\mathrm{cyc}}\sqrt{a^{2}\left(b+c\right)+2}\right)^2\cdot\sum_{\mathrm{cyc}}\left[a^2(b+c)+2\right]^2(bc+2)^3\ge \left[\sum_{\mathrm{cyc}}\left(a^2(b+c)+2\right)(bc+2)\right]^3$$and it remains to prove $$\left[\sum_{\mathrm{cyc}}\left(a^2(b+c)+2\right)(bc+2)\right]^3\ge 6(ab+bc+ca+3)\cdot\sum_{\mathrm{cyc}}\left[a^2(b+c)+2\right]^2(bc+2)^3$$which I checked it is true but not so easy.

Answer 3

By the condition, $ab+bc+ca=\frac{9-a^2-b^2-c^2}2$, so we need to prove that $$\small\sum_{\rm cyc}\sqrt{a^2(b+c)+2}\ge\sqrt{6\sum_{\rm cyc}ab+18}\Leftrightarrow\sum_{\rm cyc}-\sqrt{a^2(3-a)+2}\le-\sqrt{45-3\sum_{\rm cyc}a^2}.\tag1$$ Let $f(x)=-\sqrt{x^2(3-x)+2}$, then $$f'''(x)=\frac{3\left(\color{green}{32+144x-120x^2+40x^3}\color{blue}{+6x^5-x^6}\right)}{8\left[x^2(3-x)+2\right]^{\frac52}}>0$$ for $x\in[0,3]$. Hence, $f'(x)$ is convex for $x\in[0,3]$.

Let $a^2+b^2+c^2$ and $a+b+c$ be fixed. By EV theorem, it suffices to prove $(1)$ assuming $a=b$. In this case, $c=3-2a$, so $$(1)\Leftrightarrow2\sqrt{a^2(3-a)+2}+\sqrt{2a(3-2a)^2+2}\ge\sqrt{18+36a-18a^2}.$$ For $a\in\left[0,\frac32\right]$ we have $\sqrt{2a(3-2a)^2+2}\le\sqrt{18+36a-18a^2}$, so it suffices to prove that $$\begin{aligned}4\left[a^2(3-a)+2\right]\ge{}&2a(3-2a)^2+2+18+36a-18a^2\\&-2\sqrt{2a(3-2a)^2+2}\cdot\sqrt{18+36a-18a^2}.\end{aligned}$$ Rearranging gives $$\sqrt{2a(3-2a)^2+2}\cdot\sqrt{18+36a-18a^2}\ge6a^3-27a^2+27a+6.$$ Squaring and factorization gives $$9a(a-1)^2\left(12a^2+8-4a^3-9a\right)\ge0,$$ which is obviously true for $a\in\left[0,\frac32\right]$. We are done.