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Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

1a+1b+1c2(ab+bc+ca+2a2+b2+c2). \sqrt{1-a} + \sqrt{1-b} + \sqrt{1-c} \le \sqrt{2}(\sqrt{ab+bc+ca} + 2\sqrt{a^2+b^2+c^2}).
for non-negative a,b,ca, b, c with a+b+c=1a + b + c = 1.

Solution

From the Cauchy-Schwartz inequality for the collections (α,β,γ)(\sqrt{\alpha}, \sqrt{\beta}, \sqrt{\gamma}) and (αx,βy,γz)(\sqrt{\alpha x}, \sqrt{\beta y}, \sqrt{\gamma z}) we obtain:
(αx+βy+γz)(α+β+γ)(αx+βy+γz), (\alpha\sqrt{x} + \beta\sqrt{y} + \gamma\sqrt{z}) \le \sqrt{(\alpha + \beta + \gamma)(\alpha x + \beta y + \gamma z)},
and so
aa+b+bb+c+cc+aaa+b+bb+c+cc+a(a+b+c)(a2+ab+b2+bc+c2+ca)2(a2+b2+c2). \begin{aligned} & a\sqrt{a+b} + b\sqrt{b+c} + c\sqrt{c+a} \le \sqrt{a\sqrt{a+b} + b\sqrt{b+c} + c\sqrt{c+a}} \\ & \le \sqrt{(a+b+c)(a^2+ab+b^2+bc+c^2+ca)} \le \sqrt{2(a^2+b^2+c^2)}. \end{aligned}
similarly,
ba+b+cb+c+ac+a2(a2+b2+c2),ca+b+ab+c+bc+a2(ab+bc+ca). \begin{aligned} & b\sqrt{a+b} + c\sqrt{b+c} + a\sqrt{c+a} \le \sqrt{2(a^2+b^2+c^2)}, \\ & c\sqrt{a+b} + a\sqrt{b+c} + b\sqrt{c+a} \le \sqrt{2(ab+bc+ca)}. \end{aligned}

Adding all three inequalities, we get the required one.

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