Prove that :
√a2+b2a+b+√aba2+b2≤√2 for all positive reals a and b.
My solution:
Step 1:
Squaring both sides of the inequality:
√a2+b2a+b+√aba2+b2≤√2
(√a2+b2a+b+√aba2+b2)2≤(√2)2
a2+b2(a+b)2+2(√a2+b2a+b⋅√aba2+b2)+aba2+b2≤2
a2+b2(a+b)2+2√aba+b+aba2+b2≤2
a2+b2(a+b)2+aba2+b2+2√aba+b≤2(*)
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+b≥2√ab
2√aba+b≤1
It remains to show a2+b2(a+b)2+aba2+b2−1≤0(**).
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+b2−1
=a2+b2(a+b)2−12+aba2+b2−12
=2a2+2b2−(a+b)22(a+b)2+2ab−a2−b22(a2+b2)
=2a2+2b2−a2−2ab−b22(a+b)2−a2−2ab+b22(a2+b2)
=a2−2ab+b22(a+b)2−a2−2ab+b22(a2+b2)
=(a−b)22(a+b)2−(a−b)22(a2+b2)
=(a−b)2(12(a+b)2−12(a2+b2))
=(a−b)2(a2+b2−(a+b)22(a+b)2(a2+b2))
=(a−b)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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