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Problem Of The Week #422 June 22nd, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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Let $a,\,b,\,c$ and $d$ be four non-negative real numbers satisfying the condition

$2(ab+ac+ad+bc+bd+cd)+abc+abd+acd+bcd=16$

Prove that

$a+b+c+d\ge \dfrac{2}{3}(ab+ac+ad+bc+bd+cd)$

and determine when equality occurs.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,894
No one answered last week's POTW. (Sadface)

You can find the suggested solution as follows:
For $i= 1,\,2,\,3$, define $s_i$ as the average of the products of the i-element subsets of ${a,b,c,d}$. Then we must show

$3s_2+s_3=4\implies s_1\ge s_2$.

It suffices to prove the (unconstrained) homogeneous inequality

$3s_2^2s_1^2+s_3s_1^3\ge 4s_2^3$,

as then $3s_2+2_3=4$ will imply $(s_1-s_2)^3+3(s_1^3-s_2^3)\ge 0$.

We now recall two basic inequalities about symmetric means of non-negative real numbers. The first is Schur's inequality:

$3s_1^3+s_3\ge 4s_1s_2$,

while the second,

$s_1^2\ge s_2$ is a case of Maclaurin's inequality $s_i^{i+1}\ge s_{i+1}^i$.

These combine to prove the claim:

$3s_2^2s_1^2+s_3s_1^3\ge 3s_2^2s_1^2+\dfrac{s_2^2s_3}{s_1}\ge 4s_2^3$

Finally, for those who have only seen Schur's inequality in three variables, note that in general any inequality involving $s_1,\cdot,s_k$ which holds for $n\ge k$ variables also holds for $n+1$ variables, by replacing the variables $x_1,\cdots,x_{n+1}$ by the roots of the derivative of the polynomial $(x-x_1)\cdots (x-x_{n+1})$.
 
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