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Question

Show that b2c2+c2a2+a2b2>abc(a+b+c).

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Solution

Apply AMGM for a2b2,b2c2
a2b2+b2c22a2b2×b2c2
a2b2+b2c22b2ac eqn 1
Apply AMGM for c2b2,a2c2
c2b2+a2c22c2b2×a2c2
c2b2+a2c22c2bc eqn 2
Apply AMGM for a^{2}b^{2}, a^{2}c^{2}$
a2b2+a2c22a2b2×a2c2
a2b2+a2c22a2bc eqn 3

Adding corresponding sides of eqns 1,2,3 we get
a2b2+b2c2+a2c2abc(a+b+c)

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