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42515 posts

sqing

#1 h Apr 19, 2025, 7:52 AM

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Let $a,b,c,d\geq0, (a+b)(c+d)=2 .$ Prove that
$a^2+b^2+c^2+d^2-ac-bd \geq1$ Let $a,b,c,d\geq0, (a+2b)(c+2d)=2 .$ Prove that
$a^2+b^2+c^2+d^2-ac-bd \geq\frac{2}{5}$ Let $a,b,c,d\geq0, (a+2b)(2c+ d)=2 .$ Prove that
$a^2+b^2+c^2+d^2-ac-bd \geq\frac{3}{7}$

This post has been edited 1 time. Last edited by sqing, Apr 19, 2025, 7:54 AM

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#2 h Apr 19, 2025, 8:08 AM

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sqing wrote:

Let $a,b,c,d\geq0, (a+b)(c+d)=2 .$ Prove that
$a^2+b^2+c^2+d^2-ac-bd \geq1$

we have:
$2(a^2+b^2+c^2+d^2-ac-bd-1) =(a-c)^2 + (b-d)^2+(a^2+b^2+c^2+d^2-ac-ad-bc-bd)$ which is true by rearrangement. Equality for $a=b=c=d$ .

The other inequalities can be proved in a similar way.

This post has been edited 1 time. Last edited by GeoMorocco, Apr 19, 2025, 8:08 AM

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sqing

#3 h Apr 19, 2025, 8:09 AM

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Nice.Thanks.

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sqing

#4 h Apr 19, 2025, 8:14 AM

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Let $a,b,c$ be reals such that $a+2b+c=0$ and $a^2+b^2+c^2=3.$ Prove that
$abc\leq 1$ Let $a,b,c$ be reals such that $a+2b+c=0$ and $a^2+2b^2+c^2=3.$ Prove that
$abc\leq\frac{3\sqrt{3}}{8}$ Let $a,b,c$ be reals such that $a+4b+c=0$ and $a^2+2b^2+c^2=3.$ Prove that
$abc\leq\frac{3 }{5}\sqrt{\frac{6}{5}}$

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#5 h Apr 19, 2025, 10:38 AM

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$a+2b+c=0 \iff b= -\frac{a+c}{2}$ $a^2+b^2+c^2=3 \iff 5b^2-2ac=3$ $abc \leq 1 \iff 5b^3-3b-2 \leq 0 \iff b \leq 1 \iff |a+c| \leq 2$ because : $\frac{5}{4}(a+c)^2 \leq 3+ \frac{1}{2}(a+c)^2 \iff |a+c| \leq 2$ The other inequalities can be proved in a similar way. :-D

:-D

equality case : $(a,b,c)=(-1,1,-1)$

This post has been edited 2 times. Last edited by SunnyEvan, Apr 19, 2025, 10:45 AM

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#6 h Apr 19, 2025, 11:05 AM

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SunnyEvan wrote:

$a+2b+c=0 \iff b= -\frac{a+c}{2}$ $a^2+b^2+c^2=3 \iff 5b^2-2ac=3$ $abc \leq 1 \iff 5b^3-3b-2 \leq 0 \iff b \leq 1 \iff |a+c| \leq 2$ because : $\frac{5}{4}(a+c)^2 \leq 3+ \frac{1}{2}(a+c)^2 \iff |a+c| \leq 2$ The other inequalities can be proved in a similar way. :-D

:-D

equality case : $(a,b,c)=(-1,1,-1)$

The first one is trivial by Am-Gm. In fact, we have:
$3 = a^2+b^2+c^2 \geq 3\sqrt[3]{a^2b^2c^2}$ Therefore
$abc \leq |abc| \leq 1$

This post has been edited 1 time. Last edited by GeoMorocco, Apr 19, 2025, 11:05 AM

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#7 h Apr 19, 2025, 11:41 AM

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GeoMorocco wrote:

SunnyEvan wrote:

$a+2b+c=0 \iff b= -\frac{a+c}{2}$ $a^2+b^2+c^2=3 \iff 5b^2-2ac=3$ $abc \leq 1 \iff 5b^3-3b-2 \leq 0 \iff b \leq 1 \iff |a+c| \leq 2$ because : $\frac{5}{4}(a+c)^2 \leq 3+ \frac{1}{2}(a+c)^2 \iff |a+c| \leq 2$ The other inequalities can be proved in a similar way. :-D

:-D

equality case : $(a,b,c)=(-1,1,-1)$

The first one is trivial by Am-Gm. In fact, we have:
$3 = a^2+b^2+c^2 \geq 3\sqrt[3]{a^2b^2c^2}$ Therefore
$abc \leq |abc| \leq 1$

Yeah .

:-D

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sqing

#8 h Apr 19, 2025, 12:23 PM

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Thank you.

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