# Dan Sitaru's Inequality with Radicals and Cosines

### Problem

### Solution 1

The inequality is equivalent to

$\displaystyle (a^2+b^2+c^2)^3\ge 6^3(abc)^2\cos A\cos B\cos C.$

With the AM-GM inequality, $(a^2+b^2+c^2)^3\ge 3^3(abc)^2.$ Thus suffice it to prove that $1\ge 8\cos A\cos B\cos C.$ But it is known that

$\displaystyle \cos A+\cos B+\cos C\le \frac{3}{2}.$

So, applying the AM-GM inequality the second time, we get

$\displaystyle \frac{3}{2}\ge \cos A+\cos B+\cos C\ge 3\sqrt[3]{\cos A\cos B\cos C},$

which proves the required $1\ge 8\cos A\cos B\cos C.$

### Solution 2

Applying the Cosine rule, the inequality can be reduced to

$ (a^2+b^2+c^2)^3\geq 27(b^2+c^2-a^2)(c^2+a^2-b^2)(a^2+b^2-c^2).$

All the three bracketed terms on the right hand side are positive as the triangle is acute angled. From the AM-GM inequality,

$\displaystyle \begin{align} (a^2+b^2+c^2)&=(b^2+c^2-a^2)+(c^2+a^2-b^2)+(a^2+b^2-c^2)\\ &\geq 3[(b^2+c^2-a^2)(c^2+a^2-b^2)(a^2+b^2-c^2)]^{1/3}. \end{align}$

Cubing both sides, we get the desired inequality.

### Acknowledgment

The problem from the Romanian Mathematical Magazine has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page where one can find additional solutions.

Solution 2 is by Amit Itagi.

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