$(a^2 + b^2 + c^2)^2 \geq (ab+bc+ca)^2\geq 3abc(a+b+c) \geq 3(abc)^2$.

Why does it follows that $(ab+bc+ca)^2\geq 3abc(a+b+c)$

Because it is equivalent with $\frac 12 ( (ab-bc)^2 + (bc-ca)^2 + (ca-ab)^2 ) \geq 0$.

Or because from this well-know inequality \[ (x+y+z)^{2}\geq 3(xy+yz+zx), \] by putting $x=bc, y=ca, z=ab$ we get \[ ab+bc+ca\geq\sqrt{3abc(a+b+c)}. \]

Problem. Let $a, b, c$ three non-negative real numbers such that $a+b+c\geq abc$. Prove that \[ a^{2}+b^{2}+c^{2}\geq abc\sqrt{3}. \] Solution 1. Just like fuzzylogic said we have the followong inequalities \[ a^{2}+b^{2}+c^{2}\geq ab+bc+ca\geq\sqrt{3abc(a+b+c)}\geq abc\sqrt{3}. \] Solution 2. Let's assume by contradiction that \[ a^{2}+b^{2}+c^{2} <abc\sqrt{3}. \] By applying Cauchy-Schwarz inequality, $3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}$ and the hipothesys $a+b+c\geq abc$ we get $abc<3\sqrt{3}$. On the other hand , by Am-Gm we have \[ abc\sqrt{3}>a^{2}+b^{2}+c^{2}\geq 3\sqrt [3] {a^{2}b^{2}c^{2}}. \] We get from here $abc>3\sqrt{3}$. We have a contradiction. Solution 3. We have \[ a+b+c\geq abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 1. \] We shall prove a stronger inequality \[ ab+bc+ca\geq abc\sqrt{3}\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\sqrt{3}. \] Now, let us denote $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$ and the problems becomes: "If $x, y, z$ are three non-negative real numbers such that $xy+yz+zx\geq 1$. Then the following holds: \[ x+y+z\geq\sqrt{3}. \] But, this last problem follows immediately from this inequality \[ (x+y+z)^{2}\geq 3(xy+yz+zx). \]

$({a+b+c \over 3})^3 \geq abc$ and $a+b+c\geq abc$ $\rightarrow$ $(a+b+c)^4\geq 27a^2b^2c^2$. Also $\sqrt{{a^2+b^2+c^2\over 3}}\geq {a+b+c\over 3}$. Now the given inequlaity straightly folows from last two.

Hello. This is one of the first inequalities that I have ever proven. I think that my method is unnecessarily long. Would anyone mind giving some feedback as to how I can improve my ability in proving olympiad inequalities. Solution First assume that $a \ge b \ge c$. Then we have also that $1/bc \ge 1/ac \ge 1/ab$. Hence by Chebyshev's inequality, $\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \ge \frac{(a+b+c)(1/bc + 1/ac +1/ab)}{3} \ge \frac{abc(1/bc + 1/ac +1/ab)}{3}$ This is equivalent to $\frac{a^2 + b^2 + c^2}{a+b+c} \ge \frac{abc}{3}$ (1) Now note the following. By AM-GM, $(a^2 + b^2 + c^2)(a+b+c) \ge 9abc \Rightarrow \frac{a^2 + b^2 + c^2}{9abc} \ge \frac{1}{a+b+c}$ (2) Combining (1) and (2) yields that $\frac{(a^2 + b^2 +c^2)^2}{9abc} \ge \frac{a^2 + b^2 + c^2}{a+b+c} \ge \frac{abc}{3}$ $\Rightarrow (a^2 + b^2 +c^2)^2 \ge 3(abc)^2 \Rightarrow a^2 + b^2 +c^2 \ge \sqrt{3}abc$ EDIT: Woops. My proof can be much improved by using AM - QM instead of Chebyshev to show (1).

let $ab=\frac{1}{x},bc=\frac{1}{y},ca=\frac{1}{z}$ then $x+y+z=1$ it suffices to prove $\sum x^2\ge\sqrt{3xyz}$ or $(\sum x^2)^2\ge 3xyz(x+y+z)$ trivialized by Muirhead.

i did not found muirhead inequality anywhere.i know that it can be obatain by applying MG-MA,but not more.could you help me please?

1.Generalization Let $ a,b,c,k,r>0$ be positive real numbers with \[{a+b+c}\geq{kabc}\] and $r\leq\sqrt{3k}.$ Show that \[a^{2}+b^{2}+c^{2}\geq{rabc}.\] C.M.A,Ion Bursuc,2006

2.Generalization Let $ a,b,c,k,r>0$ be positive real numbers ,p,q,n positive integer ,$3q>n$,$pn>q$, $\left(\frac{r}{3}\right)^{3p-1}\leq\left(\frac{k}{3}\right)^{3q-n} $with \[{a+b+c}\geq{k(abc)^p}.\] Show that \[a^{n}+b^{n}+c^{n}\geq{r(abc)^q}.\] C.M.A,Ion Bursuc,2006

The following generalized inequality holds: $$\sum_{i=1}^n a_i^2\geq\sqrt[n-1]{n}\cdot\prod_{i=1}^n a_i$$For $n\geq2; a_1,\dots,a_n\ni\mathbb{R_+}$ and $\sum_{i=1}^n a_i\geq\prod_{i=1}^n a_i$ Proof: Using a known inequality that follows from CBS or Chebyshev, Rearrange or other inequalities: $\sum_{i=1}^n a_i^2\cdot n\geq\left(\sum_{i=1}^n a_i\right)^2$ we get that $\sum_{i=1}^n a_i^2\cdot n\geq\left(\sum_{i=1}^n a_i\right)^2\geq\left(\prod_{i=1}^n a_i\right)^2$ so we need $\left(\prod_{i=1}^n a_i\right)^2\geq n\sqrt[n-1]{n}\cdot\prod_{i=1}^n a_i\Leftrightarrow\prod_{i=1}^n a_i\geq n\sqrt[n-1]{n}$ so we break the problem into two parts: 1) $\prod_{i=1}^n a_i\geq n\sqrt[n-1]{n}$ We can see that inequality holds and one can check that equality holds when $a_i=\sqrt[n-1]{n},\forall i=\overline{1,n}$ 2)$\prod_{i=1}^n a_i<n\sqrt[n-1]{n}$ $$\prod_{i=1}^n a_i<n\sqrt[n-1]{n}=\sqrt[n-1]{n^n}=\left(\sqrt[n-1]{n}\right)^n\Leftrightarrow\sqrt[n]{\prod_{i=1}^n a_i}<\sqrt[n-1]{n}$$We know that $\sum_{i=1}^n a_i^2\cdot n\geq\left(\sum_{i=1}^n a_i\right)^2; \sqrt[n]{\prod_{i=1}^n a_i}<\sqrt[n-1]{n}$ and by AM-GM: $\sum_{i=1}^n a_i\geq n\cdot\sqrt[n]{\prod_{i=1}^n a_i}$ so applying these three we get $\sum_{i=1}^n a_i^2\cdot n\geq\left(\sum_{i=1}^n a_i\right)^2\geq\left(n\cdot\sqrt[n]{\prod_{i=1}^n a_i}\right)^2=n^2\cdot\sqrt[n]{\left(\prod_{i=1}^n a_i\right)^2}=n\sqrt[n-1]{n^{n-1}}\cdot\frac{\prod_{i=1}^n a_i}{\sqrt[n]{\left(\prod_{i=1}^n a_i\right)^{n-2}}}>$ $>n\cdot\sqrt[n-1]{\frac{n^{n-1}}{n^{n-2}}}\prod_{i=1}^n a_i=n\sqrt[n-1]{n}\cdot\prod_{i=1}^n a_i\Rightarrow\sum_{i=1}^n a_i^2>\sqrt[n-1]{n}\cdot\prod_{i=1}^n a_i.$ This solution might seem hard to guess, but if you firstly do the case for $n=3$ as in the problem with this method, you can understand how the general method and form of the inequality was supposed, because the number of $a_i$'s and the coefficient in front of $\prod_{i=1}^n a_i$ depend on each other and all the inequalities used in the case $n=3$ are true in general case.

Edit: WRONG SOLUTION Solution by Muirhead, it is suffices to prove that\[a^2+b^2+c^2\geq \sqrt{3}(a+b+c)\]so this is equivalent to proving that \[(a+b+c)(a^2+b^2+c^2)^2\geq 3abc(a+b+c)^2\iff \sum_{cyc}a^5+\sum_{sym}a^4b+2\sum_{sym}a^3b^2+2\sum_{cyc}a^2b^2c\geq 3\sum_{cyc}a^3bc+6\sum_{cyc}a^2b^2c\]which follows from $2\sum_{sym}a^3b^2\geq 4\sum_{cyc}a^2b^2c$ and $\sum_{cyc}a^5+\sum_{sym}a^4b\geq \sum_{cyc}a^3bc+2\sum_{cyc}a^3bc=3\sum_{cyc}a^3bc$. HERE's THE CORRECT ONE: \[a^2+b^2+c^2\ge ab+bc+ca \ge \sqrt{3abc (a+b+c)}\ge \sqrt{3} abc\]by the well-known inequality $(x+y+z)^2\ge 3(xy+yz+zx)$.

From C-S we know that $a^{2}+b^{2}+c^{2}\geq\frac{(a+b+c)^{2}}{3}.$ From $a+b+c\geq abc$ we have$\frac{(a+b+c)^{2}}{3}\geq\frac{(abc)^{2}}{3}$ $\frac{(abc)^{2}}{3}\geq \sqrt{3}abc\Longleftrightarrow abc\geq3\sqrt{3}$ from AM-GM we have $\frac{a+b+c}{3}\geq\frac{abc}{3}\geq\sqrt[3]{abc}\implies abc\geq 3\sqrt{3}.$ Thus our proof is complete.

@Above $abc\geq 3\sqrt3$ is not correct. Take $a=b=c=1$. There is a mistake in $\frac{a+b+c}{3}\geq\frac{abc}{3}\geq\sqrt[3]{abc}\implies abc\geq 3\sqrt{3}$. How did you get it?

I think it's a mistake

If $abc\geq 3\sqrt3$ \[\frac{a^2+b^2+c^2}{abc}\geq \frac{(a+b+c)^2}{3abc}\geq \frac{(abc)^2}{3abc}=\frac{abc}{3}\geq \sqrt 3\]If $abc\leq 3\sqrt 3$ \[\frac{a^2+b^2+c^2}{abc}\geq \frac{3\sqrt[3]{a^2b^2c^2}}{abc}=\frac{3}{\sqrt[3]{abc}}\geq \frac{3}{\sqrt3}=\sqrt3\]