Cho \(a^2+b^2=2\) và (a-d).(b-c)=1. C/m \(c^2+d^2-2ad-2bc-2ab\ge-2\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
Cho \(a^2+b^2=2\) và \(\left(a-d\right)\left(b-c\right)=1\). Chứng minh \(c^2+d^2-2ad-2bc-2ab\ge-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:
\(VT+4=\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ab+b^2\right)=\left(a-d\right)^2+\left(b-d\right)^2+\left(a-b\right)^2\)theo AM-GM:\(\left(a-d\right)^2+\left(b-c\right)^2\ge2\left(a-d\right)\left(b-c\right)=2\)
và \(\left(a-b\right)^2\ge0\)
do đó \(VT+4\ge2\Leftrightarrow VT\ge2\)
Dấu = xảy ra khi a=b=1;c=d=0 ...
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=a^2+b^2+c^2+d^2-2\left(a+c\right)\left(b+d\right)\)
\(VT\ge\frac{1}{4}\left(a+b+c+d\right)^2-\frac{1}{2}\left(a+b+c+d\right)^2=-\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=d=\frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Làm tạm một câu rồi đi chơi, lát làm cho.
4)
Áp dụng bất đẳng thức Cauchy-Schwarz :
\(VT\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT\ge a+b+c+\dfrac{9}{2\left(ab+bc+ca\right)}\ge\sqrt{3\left(ab+bc+ca\right)}+\dfrac{9}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{9}{2\left(ab+bc+ca\right)}\ge3\sqrt[3]{\dfrac{27}{8}}=\dfrac{9}{2}\)
Áp dụng BĐT Cauchy ta có
\(\dfrac{b^2}{a}+a\ge2b;\) \(\dfrac{c^2}{b}+b\ge2c\); \(\dfrac{a^2}{c}+c\ge2a\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2\left(ab+bc+ac\right)}\ge a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\)Ta phải chứng minh
\(a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow4\left(a+b+c\right)\left(ab+bc+ac\right)+18\ge18\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge0\)
Áp dụng BĐT Cauchy:
\(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3\)
\(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge3\left(4.3-18\right)+18=0\)=> đpcm
Lời giải:
Từ điều kiện đã cho ta có:
\(c^2+d^2-2ad-2bc-2ab+2=c^2+d^2-2ad-2bc-2ab+a^2+b^2\)
\(=(c^2-2bc+b^2)+(d^2-2ad+a^2)-2ab\)
\(=(b-c)^2+(a-d)^2-2ab\)
\(=(b-c)^2-2(b-c)(a-d)+(a-d)^2+2(b-c)(a-d)-2ab\)
\(=(b-c-a+d)^2+2-2ab\)
\(=(b-c-a+d)^2+a^2+b^2-2ab\)
\(=(b-c-a-d)^2+(a-b)^2\geq 0\)
\(\Rightarrow c^2+d^2-2ad-2bc-2ab\geq -2\) (đpcm)
Lời giải:
Từ điều kiện đã cho ta có:
\(c^2+d^2-2ad-2bc-2ab+2=c^2+d^2-2ad-2bc-2ab+a^2+b^2\)
\(=(c^2-2bc+b^2)+(d^2-2ad+a^2)-2ab\)
\(=(b-c)^2+(a-d)^2-2ab\)
\(=(b-c)^2-2(b-c)(a-d)+(a-d)^2+2(b-c)(a-d)-2ab\)
\(=(b-c-a+d)^2+2-2ab\)
\(=(b-c-a+d)^2+a^2+b^2-2ab\)
\(=(b-c-a-d)^2+(a-b)^2\geq 0\)
\(\Rightarrow c^2+d^2-2ad-2bc-2ab\geq -2\) (đpcm)