Dan Sitaru's Inequality with Radicals and Cosines


Dan  Sitaru's Inequality with Radicals and Cosines

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.


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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