Trigonometry! #32

Consider $\sec A$ and $\sec B$ . We know that AM $\ge$ GM, and the minimum value of the LHS is achieved when the equality occurs, i.e. the terms are equal.

$\sec A=\sec B \Rightarrow A=B=\frac{\pi}{6}$

$\therefore\frac{\sec A + \sec B}{2} \ge\sqrt{\sec A\sec B} \\ \sec A + \sec B \ge 2\sqrt{ \sec^{2} \frac{\pi}{6}} \\ \sec A + \sec B \ge 2\sqrt{\frac{4}{3}} \\ \boxed{\sec A + \sec B \ge \frac{4}{\sqrt{3}}}$

The minimum value is $f_{\min}=\frac{4}{\sqrt{3}}$ .

Taking $a=4,b=\sqrt{3}$ , $\boxed{a^2+b^2=19}$ .

But, the same number can also be written as $f_{\min}=\frac{4\sqrt{3}}{3}$ . Now, taking $a=4\sqrt{3},b=3$ an equally valid answer could also be $\boxed{a^2+b^2=57}$

I assumed A and B = pi/6 (hit and trial) ,this is because secant is an increasing function and cosecant is a decreasing function , thus , most probably the angles should be half of pi/6 to be minimum !

The answer is 4/(sqrt)3.

Since $\sec (x)$ is convex in the interval $(0, \dfrac {\pi}{2} )$ , we can apply Jensen's Inequality. $\dfrac {\sec A +\sec B}{2} \geq \sec {\left (\dfrac {A+B}{2} \right )} \implies \sec A + \sec B \geq \dfrac {4}{\sqrt 3}$ Thus $a=4, b=\sqrt 3, \boxed {a^2+b^2=19.}$

We can use Jensen's Inequality to easily get it

$sec(x)$ is convex on the interval in question, so by Jensen's inequality: $\frac{sec(A)+sec(B)}{2}$ $\leq$ $sec(\frac{A+B}{2})$ .

Therefore $sec(A)+sec(B)$ $\leq$ $\frac{4}{\sqrt3}$ .

Equality is achieved when $A=B$