# If $a+b+c=3,$ prove $a\sqrt{b^2+c^2+abc}+b\sqrt{a^2+c^2+abc}+c\sqrt{b^2+a^2+abc}\ge 3\sqrt{3abc}.$

Problem. Let $$a,b,c\ge 0: a+b+c=3.$$ Prove that $$a\sqrt{b^2+c^2+abc}+b\sqrt{a^2+c^2+abc}+c\sqrt{b^2+a^2+abc}\ge 3\sqrt{3abc}.$$

I've tried to use AM-GM.

Firstly, we can rewrite the OP as$$\sum_{cyc}a\sqrt{1+\frac{b^2+c^2}{abc}}\ge 3\sqrt{3}.$$ We have $$b^2+c^2\ge 2bc$$ and it leads to wrong inequality$$a\sqrt{1+\frac{2}{a}}+b\sqrt{1+\frac{2}{b}}+c\sqrt{1+\frac{2}{c}}\ge 3\sqrt{3}.$$ Also, by AM-GM $$\sum_{cyc}a\sqrt{1+\frac{b^2+c^2}{abc}}\ge 3\sqrt[3]{abc\cdot\sqrt{\prod\limits_{cyc}\dfrac{b^2+c^2+abc}{abc}}}.$$ Hence, we need to prove $$(a^2+b^2+abc)(b^2+c^2+abc)(c^2+a^2+abc)\ge 27abc.$$ I used $$a^2+b^2+abc\ge 3\sqrt[3]{a^3b^3c},$$which gave reverse $$abc\ge 1.$$

Hope to see better ideas. Thank you.

The desired inequality is written as $$\sum_{\mathrm{cyc}}\frac{a}{3}\sqrt{b^2 + c^2 + \sqrt{abc}^2} \ge \sqrt{3abc}. \tag{1}$$

We may use Jensen's inequality.

Let $$f(x, y, z) := \sqrt{x^2 + y^2 + z^2}$$. Then $$f$$ is convex. By Jensen's inequality, we have \begin{align*} \mathrm{LHS}_{(1)} &= \frac{a}{3}f(b, c, \sqrt{abc}) + \frac{b}{3}f(c, a, \sqrt{abc}) + \frac{c}{3}f(a, b, \sqrt{abc})\\ &\ge f\left(\frac{ab + bc + ca}{3}, \frac{ab + bc + ca}{3}, \sqrt{abc}\right)\\ &= \sqrt{\left(\frac{ab + bc + ca}{3}\right)^2 + \left(\frac{ab + bc + ca}{3}\right)^2 + abc}\\ &\ge \sqrt{3abc} \end{align*} where we use $$(ab + bc + ca)^2 \ge 3(a + b + c)abc$$ in the last inequality.

We are done.

• (+1). Thank you. Is is Minkowski applying ? Aug 4 at 2:33
• @TATAbox Yes. By the way, I think it is not good to change the acceptance unless a much better solution is added (Or you do not accept an answer too early, e.g. waiting 1-3 days). Michael Rozenberg's solution is nice as well. Aug 4 at 3:52
• Thank you for reminding me. Aug 4 at 3:58

Another way.

By C-S $$\sum_{cyc}a\sqrt{b^2+c^2+abc}=\frac{1}{\sqrt{3}}\sum_{cyc}a\sqrt{(1+1+1)(b^2+c^2+abc)}\geq$$ $$\geq\frac{1}{\sqrt{3}}\sum_{cyc}a(b+c+\sqrt{abc})=\frac{1}{\sqrt3}\left(2\sqrt{\sum\limits_{cyc}(a^2b^2+2a^2bc)}+3\sqrt{abc}\right)\geq$$ $$\geq\frac{1}{\sqrt3}\left(2\sqrt{\sum_{cyc}3a^2bc}+3\sqrt{abc}\right)=\frac{1}{\sqrt3}\cdot9\sqrt{abc}=3\sqrt{3abc}.$$

• (+1). Thank you, I accept your answer. Aug 3 at 11:28

By C-S $$\sum_{cyc}a\sqrt{b^2+c^2+abc}=\sqrt{\sum_{cyc}\left(2a^2b^2+a^3bc+2ab\sqrt{(b^2+c^2+abc)(a^2+c^2+abc)}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}\left(2a^2b^2+a^3bc+2ab(ab+c^2+abc)\right)}=\sqrt{\sum_{cyc}(4a^2b^2+2a^2bc+a^3bc+2a^2b^2c)}$$ and it's enough to prove that: $$\sum_{cyc}(4a^2b^2+2a^2bc+a^3bc+2a^2b^2c)\geq27abc,$$ which after homogenization gives $$\sum_{cyc}(a^3b^2+a^3c^2-a^3bc-a^2b^2c)\geq0,$$ which is true ny Muirhead.