chứng minh \(\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\) với mọi a; b; c;d
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Ta có : \(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{2}{3}\)
\(\Leftrightarrow3\left(a+b+c+d\right)^2\ge8\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)+6\left(ab+ac+ad+bc+bd+cd\right)\ge8\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)-2\left(ab+ac+ad+bc+bd+cd\right)\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(b^2-2bd+d^2\right)+\left(c^2-2cd+d^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(c-d\right)^2\ge0\) (luôn đúng)
Vậy bđt ban đầu được chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
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 :\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ac+bc+ad+bd=\left(a+b\right)\left(c+d\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
có thiếu ĐK nào k bạn ?
áp dụng BĐT cauchy :
\(\dfrac{b}{\left(a+\sqrt{b}\right)^2}+\dfrac{d}{\left(c+\sqrt{d}\right)^2}\ge2\sqrt{\dfrac{bd}{\left(a+\sqrt{b}\right)^2\left(c+\sqrt{d}\right)^2}}=\dfrac{2\sqrt{bd}}{\left(a+\sqrt{b}\right)\left(c+\sqrt{d}\right)}\)
việc còn lại cần chứng minh \(\left(a+\sqrt{b}\right)\left(c+\sqrt{d}\right)\le2\left(ac+\sqrt{bd}\right)\)(đúng theo BĐT chebyshev)(không mất tính tổng quát giả sừ \(a\le\sqrt{b};c\le\sqrt{d}\))
dấu = xảy ra khi \(a=\sqrt{b};c=\sqrt{d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x=a+b+c;y=ab+bc+ac;z=abc\)
Suy ra : \(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(\Leftrightarrow2\left(1+z\right)+\sqrt{2\left(x^2+y^2+z^2-2xz-2y+1\right)}\ge x+y+z+1\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-2xz-2y+1\right)\ge\left(x+y-z-1\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2-2xy-2xz+2x+2yz-2y-2z+1\ge0\)
\(\Leftrightarrow\left(x-y-z+1\right)^2\ge0\) (luôn đúng)
Vậy bđt ban đầu được chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a+c\right)\left(b+d\right)+2\left(ac+bd\right)\le\left(a+c\right)\left(b+d\right)+2\left(\dfrac{\left(a+c\right)^2}{4}+\dfrac{\left(b+d\right)^2}{4}\right)\\ =\dfrac{1}{2}\left(\left(a+c\right)^2+2\left(a+c\right)\left(b+d\right)+\left(b+d\right)^2\right)\\ =\dfrac{1}{2}\left(a+c+b+d\right)^2=\dfrac{1}{2}\)
XH
(a^2 + b^2 )( c^2 + d^2 ) - ( ac + bd )^2
= a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 - a^2.c^2 - b^2.d^2 - 2.ab.cd
= a^2.d^2 + b^2.c^2 + 2a.b.c.d
= (ad + bc )^2 >=0
Vậy (a^2 + b^2 )(c^2 + d^2 ) <= ( ........)