4 variables with quadrilateral sides

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4 variables with quadrilateral sides

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Source: VL

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mihaig

#1 h Apr 25, 2025, 5:11 AM • 1 Y

Y by Sadece_Threv

Let $a,b,c,d\geq0$ satisfying
$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$ Prove
$4\left(abc+abd+acd+bcd\right)\geq3\left(a+b+c+d\right)+4.$

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Quantum-Phantom

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Quantum-Phantom

#2 h Apr 25, 2025, 9:31 AM • 1 Y

Y by Sadece_Threv

Let $p=a+b+c$ , $q=ab+bc+ca$ , $r=abc$ . By the condition, $\sum\limits_{\rm cyc}\frac1{1+a}<2$ , so $2r+q>1$ . In addition, we solve out $d=\tfrac{2+p-r}{q+2r-1}$ , so the desired inequality rewrites as
$4(r+qd)\ge3(p+d)+4,$ or after plugging in and expansion,
$-2+4 q+p q-9 r-6 p r+8r^2\ge0.\tag{1}$ Let $q$ and $r$ be fixed such that $2r+q>1$ , then it suffices to prove a linear inequality of $p$ . By the $uvw$ method, we may assume $a=b$ . In this case, (1) turns into
$2 (1+b) \left[b(2b-1)^2 c^2+2b(2-3b)c+b^2+b-1\right]\ge0.$ So it suffices to show that
$f(c)=b(2b-1)^2 c^2+2b(2-3b)c+b^2+b-1\ge0.$ The discrimination of $c$ is
$\Delta_c=-4 b(b-1)^2 \left(4b^2-b-1\right).$ If $b\ge\tfrac{1+\sqrt{17}}8$ , then $\Delta_c\le0$ and we are done. If $b\le\tfrac{1+\sqrt{17}}8<\tfrac23$ , then $f(c)$ is an increasing function. Since $\frac2{1+b}+\frac1{1+c}<2$ gives $c>\tfrac{1-b}{2b}$ , then
$f(c)\ge f\left(\frac{1-b}{2b}\right)=\frac{(b+1)^2(2b-1)^2}{4b}\ge0.$ The proof is complete.

By the way, @mihaig, do you know whether this journal is still alive? And is "VL" an extremely abbreviated form of you and Vasile Cîrtoaje?

This post has been edited 1 time. Last edited by Quantum-Phantom, Apr 25, 2025, 9:40 AM

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removablesingularity

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removablesingularity

#3 h Apr 25, 2025, 12:33 PM

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The journal is still going on, although there is almost 7 years gap.

Quantum-Phantom wrote:

By the way, @mihaig, do you know whether this journal is still alive? And is "VL" an extremely abbreviated form of you and Vasile Cîrtoaje?

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mihaig

#4 h Apr 25, 2025, 4:48 PM

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VL" an extremely abbreviated form of you and Vasile Cîrtoaje?
Yes

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arqady

#5 h Apr 26, 2025, 1:20 AM

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mihaig wrote:

Let $a,b,c,d\geq0$ satisfying
$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$ Prove
$4\left(abc+abd+acd+bcd\right)\geq3\left(a+b+c+d\right)+4.$

The Vasc's EV Method helps(corollary 1.5)

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mihaig

#6 h Apr 27, 2025, 5:18 AM

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I don't think it does

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mihaig

#7 h May 2, 2025, 10:47 AM

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See, of course,
https://artofproblemsolving.com/community/c6t243f6h3558773s1_4_variables_with_quadrilateral_sides_2

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