Let a,b,ca, b, ca,b,c be positive real numbers such that a+b+c=1a2+1b2+1c2.a + b + c = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.a+b+c=a21+b21+c21.
Prove that 2(a+b+c)≥7a2b+13+7b2c+13+7c2a+13.2(a + b + c) \geq \sqrt[3]{7a^2b + 1} + \sqrt[3]{7b^2c + 1} + \sqrt[3]{7c^2a + 1}.2(a+b+c)≥37a2b+1+37b2c+1+37c2a+1.
Find all triples (a,b,c)(a,b,c)(a,b,c) for which equality holds.