Hi there! Hopefully this helps!
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Question: "Prove that a^{4} + b^{4} + c^{4} > abc(a + b + c) , where a, b, c are different positive real numbers."
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From the AM and GM inequality, we have:
a^4 + b^4 ≥ 2a^2b^2
b^4 + c^4 ≥ 2b^2c^2
c^4 + a^4 ≥ 2a^2c^2
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From adding the inequalities we have above and dividing by 2, we have:
a^4 + b^4 + c^4 ≥ a^2b^2 + b^2c^2 + c^2a^2......1
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Now we need to repeat the process of a^2b^2, b^2c^2, and c^2a^2 to get:
a^2b^2 + b^2c^2 ≥ 2b^2ac
b^2c^2 + c^2a^2 ≥ 2c^2ab
c^2a^2 + a^2b^2 ≥ 2a^2bc
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Now, we add from what we have above and divide by 2 to get:
a^2b^2 + b^2c^2 + c^2a^2 ≥ (b^2ac + c^2ab + a^2bc) or abc(b + c + a).....2
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So, from (1) and (2) it follows:
a^4 + b^4 + c^4 ≥ abc( a + b + c)
