Cho a, b, c, d>0;a+b+c+d=4. Chứng minh rằng:
\(\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge2\)
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8 tháng 4 2015
\(VT=\frac{a+b-\left(b+d\right)}{d+b}+\frac{\left(d+c\right)-\left(b+c\right)}{b+c}+\frac{\left(b+a\right)-\left(a+c\right)}{c+a}+\frac{\left(c+d\right)-\left(a+d\right)}{a+d}\)
\(VT=\frac{a+b}{d+b}-1+\frac{\left(d+c\right)}{b+c}-1+\frac{\left(b+a\right)}{c+a}-1+\frac{\left(c+d\right)}{a+d}-1\)
\(VT=\left(a+b\right).\left(\frac{1}{d+b}+\frac{1}{a+c}\right)+\left(d+c\right).\left(\frac{1}{b+c}+\frac{1}{a+d}\right)-4\)
Chứng minh đc bđt sau: Với x; y > 0 ta có \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
Áp dụng ta có: \(VT\ge\left(a+b\right).\frac{4}{d+b+a+c}+\left(d+c\right).\frac{4}{b+c+a+d}-4\ge\frac{4.\left(a+b+c+d\right)}{a+b+c+d}-4=0\)
=> ĐPCM
HD
1
19 tháng 2 2018
Cộng 4 vào vế trái nhá
\(VT+4=\left(\dfrac{a-d}{d+b}+1\right)+\left(\dfrac{d-b}{b+c}+1\right)+\left(\dfrac{b-c}{c+a}+1\right)+\left(\dfrac{c-a}{a+d}+1\right)\)
\(=\dfrac{a+b}{d+b}+\dfrac{d+c}{b+c}+\dfrac{a+b}{c+a}+\dfrac{c+d}{a+d}\)
\(=\left(a+b\right)\left(\dfrac{1}{d+b}+\dfrac{1}{c+a}\right)+\left(c+d\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)\)
\(\ge\left(a+b\right).\dfrac{4}{a+b+c+d}+\left(c+d\right).\dfrac{4}{a+b+c+d}\)
\(=\left(a+b+c+d\right).\dfrac{4}{a+b+c+d}\)\(=4\)
\(\Rightarrow VT\ge0=VP\)(Đpcm)
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ta có: \(\frac{a}{1+b^2c}=a-\frac{ab^2c}{1+b^2c}\)
mà \(1+b^2c\ge2\sqrt{1\cdot b^2\cdot c}=2b\sqrt{c}\)
\(\Rightarrow\frac{ab^2c}{1+b^2c}\le\frac{ab^2c}{2b\sqrt{c}}=\frac{ab\sqrt{c}}{2}\)
=> \(\frac{a}{1+b^2c}\ge a-\frac{ab\sqrt{c}}{2}\)
=> \(VT\ge\left(a+b+c+d\right)-\frac12\left(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\right)\)
\(VT\ge4-\frac12\left(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\right)\)
Để VT\(\ge2\) => \(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\le4\)
mà \(\sqrt{c}\le\frac{c+1}{2}\)
=> \(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\le\frac12\left(abc+bcd+cda+dab+ab+bc+cd+da\right)\)
ta có: \(ab+bc+cd+da=b\left(a+c\right)+d\left(c+a\right)=\left(a+c\right)\left(b+d\right)\)
ta có bđt: \(\left(a+c\right)\left(b+d\right)\le\left(\frac{a+c+b+d}{2}\right)^2=4\)
ta có: \(abc+bcd+cda+dab=ac\left(b+d\right)+bd\left(c+a\right)\)
mà \(ac\le\frac{\left(a+c\right)^2}{4},bd\le\frac{\left(b+d\right)^2}{4}\)
=> \(abc+bcd+cda+dab\le\frac{\left(a+c\right)^2}{4}\left(b+d\right)+\frac{\left(b+d\right)^2}{4}\left(c+a\right)\)
\(\Rightarrow abc+bcd+cda=dab\le\frac{\left(a+c\right)\left(b+d\right)}{4}\left(a+b+c+d\right)=\left(a+c\right)\left(b+d\right)\le4\)
=> \(ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}\le\frac12\left(4+4\right)=4\)
=> \(VT\ge4-\frac12\cdot4=2\)