A quaternary inequality

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A quaternary inequality

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system of equations

inequalities unsolved

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#1 h Sep 16, 2010, 4:05 AM • 2 Y

Y by Adventure10, Mango247

If $a,b,c,d > 0$
And $a + b + c + d = 4$
Prove that $\frac{1} {{{a^2}}} + \frac{1} {{{b^2}}} + \frac{1} {{{c^2}}} + \frac{1} {{{d^2}}} \geqslant {a^2} + {b^2} + {c^2} + {d^2}$

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kuing

#2 h Sep 17, 2010, 7:00 AM • 2 Y

Y by Adventure10, Mango247

westwind wrote:

If $a,b,c,d > 0$
And $a + b + c + d = 4$
Prove that $\frac{1} {{{a^2}}} + \frac{1} {{{b^2}}} + \frac{1} {{{c^2}}} + \frac{1} {{{d^2}}} \geqslant {a^2} + {b^2} + {c^2} + {d^2}$

WLOG, assume $0<a\le b\le c\le d<4$ .
if $1+\sqrt{2}<d<4$ , then we have
$\frac{1}{{a^2 }} + \frac{1}{{b^2 }} + \frac{1}{{c^2 }} + \frac{1}{{d^2 }} \ge \frac{{27}}{{\left( {4 - d} \right)^2 }} + \frac{1}{{d^2 }}$ and $a^2 + b^2 + c^2 + d^2 < \left( {4 - d} \right)^2 + d^2$ ,
so we just need to prove $f\left( d \right) = \frac{{27}}{{\left( {4 - d} \right)^2 }} + \frac{1}{{d^2 }} - \left( {4 - d} \right)^2 - d^2 \ge 0$ , but
$f'\left( d \right) = \frac{{54}}{{\left( {4 - d} \right)^3 }} - \frac{2}{{d^3 }} - 4d + 8 > \frac{{54}}{{\left( {3 - \sqrt 2 } \right)^3 }} - \frac{2}{{\left( {1 + \sqrt 2 } \right)^3 }} - 8 = \frac{8}{{343}}\left( {561 - 233\sqrt 2 } \right) > 0$
$\Rightarrow f\left( d \right) > f\left( {1 + \sqrt 2 } \right) = \frac{{27}}{{\left( {3 - \sqrt 2 } \right)^2 }} + \frac{1}{{\left( {1 + \sqrt 2 } \right)^2 }} - \left( {3 - \sqrt 2 } \right)^2 - \left( {1 + \sqrt 2 } \right)^2 = \frac{2}{{49}}\left( {130\sqrt 2 - 121} \right) > 0.$
so the inequality holds.
if $0<d\le1+\sqrt{2}$ , then $a,b,c,d \in \left( {0,1 + \sqrt 2 } \right]$ and
$\frac{1}{{a^2 }} - a^2 - 4 + 4a = \frac{{\left( {a - 1} \right)^2 \left( {1 + \sqrt 2 - a} \right)\left( {a + \sqrt 2 - 1} \right)}}{{a^2 }} \ge 0$ ,
i.e. $\frac{1}{{a^2 }} \ge a^2 + 4 - 4a$ , so we get
$\frac{1}{{a^2 }} + \frac{1}{{b^2 }} + \frac{1}{{c^2 }} + \frac{1}{{d^2 }} \ge a^2 + b^2 + c^2 + d^2 + 16 - 4\left( {a + b + c + d} \right) = a^2 + b^2 + c^2 + d^2$ .
inequality also holds. done.

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#3 h Sep 18, 2010, 12:39 PM • 1 Y

Y by Adventure10

I hope this is correct.

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kuing

#4 h Sep 22, 2010, 5:49 AM • 2 Y

Y by Adventure10, Mango247

I guess the following two inequalies also holds for the same condition
$\frac{2}{3}\left( {\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{cd}} + \frac{1}{{da}} + \frac{1}{{ac}} + \frac{1}{{bd}}} \right) \ge a^2 + b^2 + c^2 + d^2$ $\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{cd}} + \frac{1}{{da}} \ge a^2 + b^2 + c^2 + d^2$

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arqady

#5 h Aug 11, 2012, 1:56 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

I guess the following two inequalies also holds for the same condition
$\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{cd}} + \frac{1}{{da}} \ge a^2 + b^2 + c^2 + d^2$

See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=181229

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arqady

#6 h Oct 16, 2014, 4:13 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

I guess the following two inequalies also holds for the same condition
$\frac{2}{3}\left( {\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{cd}} + \frac{1}{{da}} + \frac{1}{{ac}} + \frac{1}{{bd}}} \right) \ge a^2 + b^2 + c^2 + d^2$

I agree with you. It's true and follows from previous Vasc's inequality.

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