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in Linear Inequations by (23.5k points)
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If a, b, c are positive real numbers, then show that (a + 1)7 (b + 1)7 (c + 1)7 > 77 a4b4c4.

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LHS = (a + 1)7 (b + 1)7 (c + 1)7 

= [(a + 1) (b + 1) (c + 1)]7 

= [1 + a + b + c + ab + bc + ca + abc]7 > (a + b + c + ab + bc + ca + abc)7       …(i) 

Now using the AM, GM inequality, i.e., AM > GM, we have 

\(\frac{a+b+c+ab+bc+ca+abc}{7}\) > (a.b.c.ab.bc.ca.abc)\(\frac17\) 

\(\frac{1}{7^7}\)(a + b + c + ab + bc + ca + abc)7 (a4 b4 c4)         [Raising both sides to power 7] 

⇒ (a + b + c + ab + bc + ca + abc)7 > 77 (a4 b4 c4)      …(ii) 

From (i) and (ii) 

(a + 1)7 (b + 1)7 (c + 1)7 > 77 (a4 b4 c4).

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