Show that if a,b and c are the lengths of the sides of a right triangle with hypotenuse c, then
(a) c√ab≥√2
(b) (c−a)(c−b)(c+a)(c+b)≤17−12√2.
My solution:
(a)
c2=a²+b²≥2ab(by the AM-GM inequality)
c2ab≥2
Taking the square root on both sides completes the proof, i.e. c√ab≥√2, equality occurs when a=b.
(b)
We have c2−a2=b2⟹c−a=b2c+a.
By the similar token, we also have c−b=a2c+b, if we're going to replace these two into the original LHS of the inequality, we get:
(c−a)(c−b)(c+a)(c+b)=(b2)(a2)(c+a)2(c+b)2=(ab(c+a)(c+b))2=(abc2+c(a+b)+ab)2=(1c2ab+c(a+b)ab+1)2≤(1c2ab+c(2√ab)ab+1)2(by the AM-GM inequality)=(1(c√ab)2+2c√ab+1)2≤(1(√2)2+2√2+1)2=(3−2√2)2=17−12√2Q.E.D.
Equality occurs when a=b.