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sqing   7
N Apr 19, 2025 by sqing
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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
$$  a^2+b^2+c^2+d^2-ac-bd \geq\frac{3}{7} $$
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sqing
Apr 19, 2025
sqing
Apr 19, 2025
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sqing
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#1
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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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GeoMorocco
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#2
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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
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#3
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Nice.Thanks.
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sqing
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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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SunnyEvan
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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
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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GeoMorocco
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#6
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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
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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SunnyEvan
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#7
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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
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
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#8
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Thank you.
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