Inequality with a,b,c,d

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Source: Moroccan Training 2025

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#1 h Apr 9, 2025, 1:35 PM

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Let $a,b,c,d$ positive real numbers such that $a+b+c+d=3+\frac{1}{abcd}$ . Prove that :
$a^2+b^2+c^2+d^2+5abcd \geq 9$

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42555 posts

sqing

#3 h Apr 9, 2025, 1:55 PM

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Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=3+\frac{1}{abcd}$ . Prove that
$a^2+b^2+c^2+d^2+\frac{11}{2}abcd \geq \frac{19}{2}$

This post has been edited 1 time. Last edited by sqing, Apr 9, 2025, 1:59 PM

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sqing

#4 h Apr 10, 2025, 12:05 PM

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Let $a,b,c,d$ positive real numbers such that $a+b+c+d=3+\frac{1}{abcd}$ . Prove that
$a^3+b^3+c^3+d^3+3+abcd \geq 2(a^2+b^2+c^2+d^2)$

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#5 h Apr 18, 2025, 12:19 PM

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any solutions???

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arqady

#6 h Apr 18, 2025, 5:10 PM • 1 Y

Y by teomihai

GeoMorocco wrote:

Let $a,b,c,d$ positive real numbers such that $a+b+c+d=3+\frac{1}{abcd}$ . Prove that :
$a^2+b^2+c^2+d^2+5abcd \geq 9$

We can show that for any $k\leq5\frac{5}{9}$ , $a>0$ , $b>0$ , $c>0$ and $d>0$ such that: $a+b+c+d=3+\frac{1}{abcd}$ the following inequality holds. $a^2+b^2+c^2+d^2+kabcd \geq4+k.$

This post has been edited 4 times. Last edited by arqady, Apr 18, 2025, 7:25 PM

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#7 h Apr 18, 2025, 9:20 PM

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arqady wrote:

GeoMorocco wrote:

Let $a,b,c,d$ positive real numbers such that $a+b+c+d=3+\frac{1}{abcd}$ . Prove that :
$a^2+b^2+c^2+d^2+5abcd \geq 9$

We can show that for any $k\leq5\frac{5}{9}$ , $a>0$ , $b>0$ , $c>0$ and $d>0$ such that: $a+b+c+d=3+\frac{1}{abcd}$ the following inequality holds. $a^2+b^2+c^2+d^2+kabcd \geq4+k.$

What is your solution for $k=5$ arqady?

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arqady

#8 h Apr 19, 2025, 2:30 AM

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GeoMorocco wrote:

What is your solution for $k=5$ arqady?

I used the Vasc's EV Method: It's enough to assume that $b=c=d.$
Also, LM helps.

This post has been edited 1 time. Last edited by arqady, Apr 19, 2025, 2:32 AM

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