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5 Feb 2010 23:13:34 IST

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If a,b,c are positive real nos; PROVE (a+1)^7(b+1)^7(c+1)^7 >7^7 (abc)^4

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If a,b,c are positive real nos; PROVE (a+1)^7(b+1)^7(c+1)^7 >7^7 (abc)^4

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Hari Shankar

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Joined: 28 Feb 2007

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6 Feb 2010 06:59:48 IST

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Equivalently we have to prove that

$\left(a^{\frac{3}{7}} + \frac{1}{a^{\frac{4}{7}}}\right)^7\left(b^{\frac{3}{7}} + \frac{1}{b^{\frac{4}{7}}}\right)^7\left(c^{\frac{3}{7}} + \frac{1}{c^{\frac{4}{7}}}\right)^7>7^7$

From AM-GM Inequality we have

$\left(\frac{a^{\frac{3}{7}}}{4} + \frac{a^{\frac{3}{7}}}{4} +\frac{a^{\frac{3}{7}}}{4} +\frac{a^{\frac{3}{7}}}{4} +\frac{1}{3a^{\frac{4}{7}}} +\frac{1}{3a^{\frac{4}{7}}} +\frac{1}{3a^{\frac{4}{7}}} \right)^7 > \frac{7^7}{4^4 3^3}$

So we have to prove that

$\left( \frac{7^7}{4^4 3^3} \right)^3>7^7$ or that $7^{14}>4^{12} 3^9$

From Jensen's Inequality, we have $\frac{12 \log 4 + 9 \log 3}{21} < \log \left( \frac{75}{21} \right) \Rightarrow 4^{12}3^9 < \left(\frac{25}{7} \right)^{21}$

So we are left to prove that $7^{14}> \left(\frac{25}{7} \right)^{21}$ or that 7^5>25^3 which is easy to prove

Bipin Dubey

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Joined: 23 Jan 2007

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17 Mar 2010 20:58:46 IST

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This was my JEE-2004 question. I solved in this way :

(AM - GM inequality)

Add 1 to the numerator of LHS and drop the equality sign

Transmigrator

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Joined: 21 May 2008

Posts: 426

17 Mar 2010 21:09:13 IST

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Hats off to u Bipin Sir.................You make Maths look so simple, easy and elegant............No high-level theorems, no long steps................Wow!!

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