Question

Show that, If a not equal to b .If a+b=1, then
$({a + \frac{1}{a} })^{2} + ({b + \frac{1}{b} })^{2} > \frac{25}{2}$
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Answer 1

Answer:

If a,ba,b are positive real numbers and a+b=1a+b=1, prove that :

(a+1a)2+(b+1b)2≥252

(a+1a)2+(b+1b)2≥252

I can see that the value 252252 is attained for a=b=12a=b=12. But I do not know how to show that this is the minimal possible value.

Thank you.