Inequalities Question

**Chandru1** · June 20th 2009, 09:49 PM

If $a,b,c$ are positive real numbers prove that $\Bigl[ (1+a)(1+b)(1+c) \Bigr]^{7}>7^{7} a^{4}b^{4}c^{4}$

**simplependulum** · June 20th 2009, 10:08 PM

Use AM $\geq$ GM . $a^7 + b^7 + c^7 + (ab)^7 + (ac)^7 + (bc)^7 + (abc)^7 \geq 7 (abc)^4$

Replace a,b,c by $a^{\frac{1}{7}}, b^{\frac{1}{7}}, c^{\frac{1}{7}}$
it becomes :

$a + b + c + ab + ac + bc + abc \geq 7 (abc)^{\frac{4}{7}}$

Therefore ,

$1 + a + b + c + ab + ac + bc + abc \geq 1 + 7 (abc)^{\frac{4}{7}} > 7 (abc)^{\frac{4}{7}}$

Then ,

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

$[(1+a)(1+b)(1+c)]^7 > 7^7 (abc)^4$

**NonCommAlg** · June 21st 2009, 01:09 AM

Originally Posted by simplependulum

Use AM $\geq$ GM . $a^7 + b^7 + c^7 + (ab)^7 + (ac)^7 + (bc)^7 + (abc)^7 \geq 7 (abc)^4$

Replace a,b,c by $a^{\frac{1}{7}}, b^{\frac{1}{7}}, c^{\frac{1}{7}}$
it becomes :

$a + b + c + ab + ac + bc + abc \geq 7 (abc)^{\frac{4}{7}}$

Therefore ,

$1 + a + b + c + ab + ac + bc + abc \geq 1 + 7 (abc)^{\frac{4}{7}} > 7 (abc)^{\frac{4}{7}}$

Then ,

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

$[(1+a)(1+b)(1+c)]^7 > 7^7 (abc)^4$

similarly, for any positive real numbers $x_1,x_2, \cdots , x_n$ we have: $(1+x_1)(1+x_2) \cdots (1+x_n) \geq 1 + (2^n - 1) \sqrt[2^n - 1]{(x_1x_2 \cdots x_n)^{2^{n-1}}}.$ as a result:

$[(1+x_1)(1+x_2) \cdots (1+x_n)]^{2^n - 1} > (2^n - 1)^{2^n - 1} (x_1x_2 \cdots x_n)^{2^{n-1}}.$

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