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Sunday, September 18, 2016

If a,b and c are the lengths of the sides of a right triangle with hypotenuse c, prove (ca)(cb)(c+a)(c+b)17122.

Show that if a,b and c are the lengths of the sides of a right triangle with hypotenuse c, then
(a) cab2
(b) (ca)(cb)(c+a)(c+b)17122.

My solution:

(a)

c2=a²+b²2ab(by the AM-GM inequality)

c2ab2
Taking the square root on both sides completes the proof, i.e. cab2, equality occurs when a=b.

(b)

We have c2a2=b2ca=b2c+a.

By the similar token, we also have cb=a2c+b, if we're going to replace these two into the original LHS of the inequality, we get:

(ca)(cb)(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(2ab)ab+1)2(by the AM-GM inequality)=(1(cab)2+2cab+1)2(1(2)2+22+1)2=(322)2=17122Q.E.D.

Equality occurs when a=b.