Chứng minh bất đẳng thức mincopxki

mincopxki mik chưa nghe qua

thì đó bài bạn đó, BĐT ở bài bạn chính là Mincopxki, bn click vào link đó có cách chứng minh đó :V

Ta chứng minh: \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\left(a+c\right)^2+\left(b+d\right)^2\)

\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge ac+bd\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)

\(\Leftrightarrow a^2b^2-2abcd+c^2d^2=\left(ab-cd\right)^2\ge0\)(luôn đúng)

Tương tự cho \(\sqrt{\left(a+c\right)^2+\left(b+d\right)}^2,\sqrt{m^2+n^2}\), chứng minh được:

\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{m^2+n^2}\ge\sqrt{\left(a+c+n\right)^2}+\sqrt{\left(b+d+n\right)^2}\)(BDT Minkowski)

\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\ge\left(a+c\right)^2+\left(b+d\right)^2\)

\(\Leftrightarrow2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\ge2ac+2bd\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)

BĐT cuối đúng theo BĐT Bunhiacopski 

Dấu "=" khi \(\frac{a}{c}=\frac{b}{d}\)

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 :\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ac+bc+ad+bd=\left(a+b\right)\left(c+d\right)\)

Áp dụng BĐT Bunhiacopxki:

(�2+�2)(�2+�2)≥(��+��)2=��+��(a2+c2)(b2+c2)​≥(ac+bc)2​=ac+bc

CMTT : (�2+�2)(�2+�2)≥��+��(a2+d2)(b2+d2)​≥ad+bd

Ta có :(�2+�2)(�2+�2)+(�2+�2)(�2+�2)≥��+��+��+��=(�+�)(�+�)(a2+c2)(b2+c2)​+(a2+d2)(b2+d2)​≥ac+bc+ad+bd=(a+b)(c+d)

\(\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}\)

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 ) 

.

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

Bđt Bu-nhia-cop-xki \(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\), đẳng thức xảy ra khi \(ay=bx\)

a.

\(\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\le\left(2+3\right)\left(2x^2+3y^2\right)=5^2\)

\(\Rightarrow-5\le2x+3y\le5\)

b.

\(\sqrt{a+c}.\sqrt{b+c}+\sqrt{a-c}.\sqrt{b-c}\le\sqrt{a+c+a-c}.\sqrt{b+c+b-c}\)

\(=\sqrt{2a}.\sqrt{2b}=2\sqrt{ab}\)

Dấu bằng xảy ra khi \(\frac{\sqrt{a+c}}{\sqrt{a-c}}=\frac{\sqrt{b+c}}{\sqrt{b-c}}\), hay \(a=b\)

Thử lại với a = b thì \(VT=2a=2\sqrt{ab}=VP>\sqrt{ab}\) nên đề đã ra sai vế phải của bđt.

c.

bđt \(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)

d.

bđt \(\Leftrightarrow\left(a+c\right)^2+\left(b+d\right)^2\le a^2+b^2+c^2+d^2+2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\)

\(\Leftrightarrow ac+bd\le\sqrt{a^2+b^2}.\sqrt{c^2+d^2}\)

bđt trên luôn đúng vì theo bđt Bu-nhia-cop-xki, ta có:

\(\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\sqrt{\left(ac+bd\right)^2}=\left|ac+bd\right|\ge ac+bd\)