Friday, May 6, 2016

Prove that : a2+b2a+b+aba2+b22 for all positive reals a and b.

Prove that :
a2+b2a+b+aba2+b22 for all positive reals a and b.

My solution:

Step 1:

Squaring both sides of the inequality:

a2+b2a+b+aba2+b22

(a2+b2a+b+aba2+b2)2(2)2

a2+b2(a+b)2+2(a2+b2a+baba2+b2)+aba2+b22

a2+b2(a+b)2+2aba+b+aba2+b22

a2+b2(a+b)2+aba2+b2+2aba+b2(*)

Up to this point, if we can prove the above inequality (*) as correct, then we're done.

Step 2:

Observe that the last term on the LHS of the inequality (*) above is less than or equal to 1, since:

a+b2ab

2aba+b1

It remains to show a2+b2(a+b)2+aba2+b210(**).

Step 3:

If we distribute the 1 as two negative one half and give it to the two terms on the LHS of (**), we see that we have:

a2+b2(a+b)2+aba2+b21

=a2+b2(a+b)212+aba2+b212

=2a2+2b2(a+b)22(a+b)2+2aba2b22(a2+b2)

=2a2+2b2a22abb22(a+b)2a22ab+b22(a2+b2)

=a22ab+b22(a+b)2a22ab+b22(a2+b2)

=(ab)22(a+b)2(ab)22(a2+b2)

=(ab)2(12(a+b)212(a2+b2))

=(ab)2(a2+b2(a+b)22(a+b)2(a2+b2))

=(ab)2(2ab2(a+b)2(a2+b2))

0 since both a and b are positive real numbers and the inequality follows.


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