Solution
This is equivalent to: (a+b+c)^2 + 3(ab+bc+ca) \geq 6(a+b+c)(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) which is the cyclic sum of: (a+b+c)^2 + 9ab \geq 6 \sqrt{ab}(a+b+c) which is true by AM-GM.
Generalization
Find all \lambda such that this holds for all a,b,c \geq 0: (a+b+c)^2 + \lambda (ab+bc+ca) \geq \frac{\lambda + 3}{9}(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2
Solution
For 0 \leq \lambda \leq 3/2 we can show it by proving the original problem, and realizing that: 3(a+b+c)^2 \geq (a+b+c)(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 is just AM-RMS.
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