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dễ thôi
ta có:
\(\frac{a}{1+b^2c}=a-\frac{ab^2c}{1+b^2c};\frac{b}{1+c^2d}=b-\frac{bc^2d}{1+c^2d};\frac{c}{1+d^2a}=c-\frac{cd^2a}{1+d^2a};\frac{d}{1+a^2b}=d-\frac{da^2b}{1+a^2b}\)
áp dụng cauchy ta có:
\(b^2c+1\ge2b\sqrt{c};c^2d+1\ge2c\sqrt{d};d^2a+1\ge2d\sqrt{a};a^2b+1\ge2a\sqrt{b}\)
\(=4-\frac{ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}}{2}\)
theo ông cauchy thì
\(ab\sqrt{c}\le\frac{ab\left(c+1\right)}{2};bc\sqrt{d}\le\frac{bc\left(d+1\right)}{2};cd\sqrt{a}\le\frac{cd\left(a+1\right)}{2};da\sqrt{b}\le\frac{da\left(b+1\right)}{2}\)
\(\Rightarrow4-\frac{ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+da\sqrt{b}}{2}\ge4-\frac{\left(abc+bcd+cda+dab\right)+\left(ab+bc+cd+da\right)}{4}\)
vẫn là ông cauchy nói là \(abc+bcd+cda+dab\le\frac{1}{16}\left(a+b+c+d\right)^3=4\)
\(ab+bc+cd+da=\left(b+d\right)\left(a+c\right)\le\frac{\left(a+b+c+d\right)^2}{4}=4\)
\(\Rightarrow4-\frac{\left(abc+bcd+cda+dab\right)+\left(ab+bc+cd+da\right)}{4}\ge4-\frac{4+4}{4}=2\)
\(\Rightarrow\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge2\left(Q.E.D\right)\)
dấu bằng xảy ra khi a=b=c=d=1
\(\Rightarrow\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge\left(a+b+c+d\right)-\frac{ab^2c}{2b\sqrt{c}}-\frac{bc^2d}{2c\sqrt{d}}-\frac{cd^2a}{2d\sqrt{a}}-\frac{da^2b}{2a\sqrt{b}}\)
Giải:
Áp dụng BĐT AM - GM ta có:
\(\dfrac{a}{1+b^2c}=a-\dfrac{ab^2c}{1+b^2c}\ge a-\dfrac{ab^2c}{2b\sqrt{c}}\) \(=a-\dfrac{ab\sqrt{c}}{2}\)
\(\ge a-\dfrac{b\sqrt{a.ac}}{2}\ge a-\dfrac{b\left(a+ac\right)}{4}\) \(\ge a-\dfrac{1}{4}\left(ab+abc\right)\)
\(\Rightarrow\dfrac{a}{1+b^2c}\ge a-\dfrac{1}{4}\left(ab+abc\right).\) Tượng tự ta cũng có:
\(\dfrac{b}{1+c^2d}\ge b-\dfrac{1}{4}\left(bc+bcd\right);\dfrac{c}{1+d^2a}\ge c-\dfrac{1}{4}\left(cd+cda\right);\dfrac{d}{1+a^2b}\ge d-\dfrac{1}{4}\left(da+dab\right)\)
Cộng theo vế 4 BĐT trên ta được:
\(\dfrac{a}{1+b^2c}+\dfrac{b}{1+c^2d}+\dfrac{c}{1+d^2a}+\dfrac{d}{1+a^2b}\)
\(\ge a+b+c+d-\dfrac{1}{4}\)\(\left(ab+bc+cd+da+abc+bcd+cda+dab\right)\)
Lại áp dụng BĐT AM - GM ta có:
\(ab+bc+cd+da\) \(\le\dfrac{1}{4}\left(a+b+c+d\right)^2=4\)
\(abc+bcd+cda+dab\) \(\le\dfrac{1}{16}\left(a+b+c+d\right)^3=4\)
Do đó:
\(\dfrac{a}{1+b^2c}+\dfrac{b}{1+c^2d}+\dfrac{c}{1+d^2a}+\dfrac{d}{1+a^2b}\)
\(\ge a+b+c+d-2=2\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=c=d=1\)
\(N=\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\)
Áp dụng BĐT Cauchy ta có:
\(\frac{a}{1+b^2c}=a-\frac{ab^2c}{1+b^2c}\)
\(\ge a-\frac{ab^2c}{2b\sqrt{c}}=a-\frac{ab\sqrt{c}}{2}=a-\frac{b\sqrt{ac}\sqrt{a}}{2}\)
\(\ge a-\frac{b\left(ac+c\right)}{4}\).Suy ra \(\frac{a}{1+b^2c}\ge a-\frac{1}{4}\cdot\left(ab+abc\right)\)
Tương tự ta có:
\(\frac{b}{a+c^2d}\ge b-\frac{1}{4}\left(bc+bcd\right)\)
\(\frac{c}{1+d^2a}\ge c-\frac{1}{4}\left(cd+cda\right)\)
\(\frac{d}{1+a^2b}\ge d-\frac{1}{4}\left(da+dab\right)\)
Do đó: \(S=\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\)
\(\ge a+b+c+d-\frac{1}{4}\left(ab+bc+cd+da+abc+bcd+cda+dab\right)\)
\(=4-\frac{1}{4}\left(ab+bc+cd+da+abc+bcd+cda+dab\right)\)
Ta có:
\(ab+bc+cd+da\le\frac{1}{4}\left(a+b+c+d\right)^2=4\)
\(abc+bcd+cda+dab\le\frac{1}{16}\left(a+b+c+d\right)^3=4\)
nên \(S\ge4-\frac{1}{4}\cdot\left(4+4\right)=2\)(Đpcm)
Dấu = khi \(a=b=c=d=1\)
Bài 1:
\(P=(x+1)\left(1+\frac{1}{y}\right)+(y+1)\left(1+\frac{1}{x}\right)\)
\(=2+x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}\)
Áp dụng BĐT Cô-si:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(x+\frac{1}{2x}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
\(y+\frac{1}{2y}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
Áp dụng BĐT SVac-xơ kết hợp với Cô-si:
\(\frac{1}{2x}+\frac{1}{2y}\geq \frac{4}{2x+2y}=\frac{2}{x+y}\geq \frac{2}{\sqrt{2(x^2+y^2)}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Cộng các BĐT trên :
\(\Rightarrow P\geq 2+2+\sqrt{2}+\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
Vậy \(P_{\min}=4+3\sqrt{2}\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)
Bài 2:
Áp dụng BĐT Svac-xơ:
\(\frac{1}{a+3b}+\frac{1}{b+a+2c}\geq \frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
\(\frac{1}{b+3c}+\frac{1}{b+c+2a}\geq \frac{4}{2b+4c+2a}=\frac{2}{b+2c+a}\)
\(\frac{1}{c+3a}+\frac{1}{c+a+2b}\geq \frac{4}{2c+4a+2b}=\frac{2}{c+2a+b}\)
Cộng theo vế và rút gọn :
\(\Rightarrow \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\geq \frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$