Inequality with 4 acute angles

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Inequality with 4 acute angles

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#1 h Oct 23, 2010, 5:30 PM • 1 Y

Y by Adventure10

FOr non zero positive angles $A, B,C, D$ each less than $\frac{\pi}{2}$ , and $A + B + C + D = \pi$ ;
prove that $\sqrt{2}(\tan A + \tan B + \tan C + \tan D ) \ge \sec A + \sec B + \sec C + \sec D$

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kuing

#2 h Oct 24, 2010, 3:58 AM • 2 Y

Y by Adventure10, Mango247

Agr_94_Math wrote:

FOr non zero positive angles $A, B,C, D$ each less than $\frac{\pi}{2}$ , and $A + B + C + D = \pi$ ;
prove that $\sqrt{2}(\tan A + \tan B + \tan C + \tan D ) \ge \sec A + \sec B + \sec C + \sec D$

let $a=\tan A,b=\tan B,c=\tan C,d=\tan D$ , ineq becomes

$\sqrt2 (a+b+c+d)\ge\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}$ .

from $A + B + C + D = \pi$ we can prove that $a+b+c+d=abc+bcd+cda+dab$ , then we can get $a^2+1=\frac{(a+b)(a+c)(a+d)}{a+b+c+d}$ and so on, so ineq becomes

$\sqrt{2(a+b+c+d)^3} \ge \sum{\sqrt{(a+b)(a+c)(a+d)}}$

or

$2(a+b+c+d)^3 \ge \left(\sum{\sqrt{(a+b)(a+c)(a+d)}}\right)^2$

to be continue...

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#3 h Oct 24, 2010, 4:38 AM • 1 Y

Y by Adventure10

kuing wrote:

to be continue...

how? Is it obvious ?

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kuing

#4 h Oct 24, 2010, 5:19 AM • 1 Y

Y by Adventure10

continue:

by Cauchy-Schwarz Inequality, we have

$\begin{align*} &\left( {\sum_{cyc} {\sqrt {(a + b)(a + c)(a + d)} } } \right)^2 \\ \le &\left( {\sum_{cyc} {(a + c)(a + d)} } \right)\left( {\sum_{cyc} {(a + b)} } \right) \\ = &\left( {\sum_{cyc} {a^2}+{2(ab + ac + ad + bc + bd + cd)} } \right)\left( {2(a + b + c + d)} \right) \\ = &2(a + b + c + d)^3 \end{align*}$

done.

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#5 h Oct 24, 2010, 5:55 AM • 2 Y

Y by Adventure10, Mango247

Congrats! and nice proof kuing!

I tried same like you, but I did not deduce this -
$a^{2}+1=\frac{(a+b)(a+c)(a+d)}{a+b+c+d}$

It was asked in a local maths contest.

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