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

71
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)

21
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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