Trigonometric Inequality

\hspace{-16}$If $\mathbf{\alpha\in \mathbb{R}}\;,$ Then Prove That\\\\\\ $\mathbf{\sqrt{17}\leq\sqrt{\cos^2 \alpha+4\cos \alpha +6}+\sqrt{\cos^2 \alpha-2\cos \alpha +3}\leq \sqrt{2}+\sqrt{11}}$

2 Answers

Vivek @ Born this Way ·

We write the terms inside the root as perfect squares. Now let us define a function :

f(α) = (cos α+2)2+(cos α -1)2 and find the minimum and maximum values of this function.

This maximum and minimum values will correspond to the given inequality. (I mean when put into the real expression)

Shubhodip ·

Consider f(x)= (x^2 + 4x+ 6)^(1/2) + (x^2 -2x+ 3)^(1/2)

It's convex when -1≤x≤1

so f is max either at x= 1 or -1

we see its 1, so RHS proved...

f'(1/2) = 0, since f is convex in [-1,1], f(1/2) gives the minimum value in [-1,1] which is √17 so LHS proved....

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