Cm\(\sqrt{\left(A^2+C^2\right)\left(B^2+D^2\right)}+\sqrt{\left(A^2+D^2\right)\left(B^2+C^2\right)}\)
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A
22 tháng 5 2023
Áp dụng BĐT Bunhiacopxki:
\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\ge\sqrt{\left(ac+bc\right)^2}=ac+bc\)
CMTT : \(\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ad+bd\)
Ta có :\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ac+bc+ad+bd=\left(a+b\right)\left(c+d\right)\)
19 tháng 6 2019
\(\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}\)
Cần CM : \(\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}\ge\left|a+b\right|-\left|c+d\right|\)
\(\Leftrightarrow\)\(\left(a+b\right)^2+\left(c+d\right)^2\ge\left(a+b\right)^2+\left(c+d\right)^2-2\left|\left(a+b\right)\left(c+d\right)\right|\)
\(\Leftrightarrow\)\(\left|\left(a+b\right)\left(c+d\right)\right|\ge0\) ( luôn đúng \(\forall\left|a+b\right|\ge\left|c+d\right|\) )
Do đó \(VT\ge\left|a+b\right|-\left|c+d\right|=\left(\sqrt{\left|a+b\right|}\right)^2-\left(\sqrt{\left|c+d\right|}\right)^2\)
\(=\left(\sqrt{\left|a+b\right|}+\sqrt{\left|c+d\right|}\right)\left(\sqrt{\left|a+b\right|}-\sqrt{\left|c+d\right|}\right)\)
\(\ge2\sqrt[4]{\left|a+b\right|.\left|c+d\right|}\left(\sqrt{\left|a+b\right|}-\sqrt{\left|c+d\right|}\right)\)
\(=2\left(\sqrt[4]{\left|a+b\right|^3.\left|c+d\right|}-\sqrt[4]{\left|a+b\right|.\left|c+d\right|^3}\right)\) ( đpcm )
.
19 tháng 6 2019
Áp dụng bất đẳng thức Mincoxki ta có
\(\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}\)
Buniacoxki \(\sqrt{\left(\left(a+b\right)^2+\left(c+d\right)^2\right)\left(1+1\right)}\ge|a+b|+|c+d|\)
Khi đó cần Cm
\(|a+b|+|c+d|\ge2\left(\sqrt{|a+b|^3|c+d|}-\sqrt{|c+d|^3|a+b|}\right)\)
Đặt \(\sqrt[4]{|a+b|}=x,\sqrt[4]{|c+d|}=y\left(x,y\ge0\right)\)
Cần Cm \(x^4+y^4\ge2\left(x^3y-xy^3\right)\left(1\right)\)
<=> \(x^3\left(x-2y\right)+y^4+2xy^3\ge0\left(2\right)\)
+ Nếu \(x\ge2y\)=> BĐT được CM
+ Nếu \(x\le2y\)
(1) <=> \(x^4+y^4+2xy^3\ge2x^3y\)
Mà \(x^4+x^2y^2\ge2x^3y\)
=> Cần CM \(y^4+2xy^3-x^2y^2\ge0\)
<=> \(y^4+xy^2\left(2y-x\right)\ge0\)luôn đúng do \(x\le2y\)
=> BĐT được CM
Dấu bằng xảy ra khi a=b=c=d=0
4 tháng 11 2019
\(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Đây nè @Võ Hồng Phúc(Phúc bím)
20 tháng 5 2022
a: \(=3\sqrt{5}-\left(\sqrt{5}-2\right)=2\sqrt{5}+2\)
b: \(=\left|a-b\right|-\left|b-c\right|-\left|c-d\right|\)
\(=b-a-\left(c-b\right)-\left(d-c\right)\)
=b-a-c+b-d+c
=2b-d-a
26 tháng 6 2021
\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)
\(=54+8-32=30\)
\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)
\(=5-2\sqrt{2}\)
\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)
\(=2-2\sqrt{3}\)
\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)
\(=2\sqrt{6}\)
\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)
26 tháng 6 2021
`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`
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