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\(BDT\Leftrightarrow\sqrt[3]{\frac{abc}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}+\sqrt[3]{\frac{xyz}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}\le1\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt[3]{\frac{abc}{(a+x)(b+y)(c+z)}}\le\frac{\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}}{3}\)
\(\sqrt[3]{\frac{xyz}{(a+x)(b+y)(c+z)}}\le\frac{\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}}{3}\)
\(\Rightarrow VT\le\frac{\frac{x+a}{x+a}+\frac{b+y}{b+y}+\frac{c+z}{c+z}}{3}=1\)
Xảy ra khi a=b=c và x=y=z
Áp dụng BĐT AM-Gm:
\(\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}\ge3\sqrt[3]{\frac{abc}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}\)
\(\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}\ge3\sqrt[3]{\frac{xyz}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}\)
Cộng 2 BĐT trên theo vế:
\(3\ge3.\frac{\sqrt[3]{abc}+\sqrt[3]{xyz}}{\sqrt[3]{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}\)
\(\Leftrightarrow\sqrt[3]{\left(a+x\right)\left(b+y\right)\left(c+z\right)}\ge\sqrt[3]{abc}+\sqrt[3]{xyz}\)(đpcm)
Dấu = xảy ra khi \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
)
Đặt \(\sqrt{1+a^2}+\sqrt{1-a^2}=x\Rightarrow\sqrt{2}\le x\le2\)
\(x^2=2+2\sqrt{1-a^4}\Rightarrow\sqrt{1-a^4}=\dfrac{x^2-2}{2}\)
\(\Rightarrow\dfrac{x^2-2}{2}+\left(b+1\right)x+b-4\le0\)
\(\Rightarrow x^2+2\left(b+1\right)x+2b-10\le0\)
\(\Rightarrow x^2+2x-10\le-2b\left(x+1\right)\)
\(\Rightarrow-2b\ge\dfrac{x^2+2x-10}{x+1}\)
\(\Rightarrow-2b\ge\max\limits_{\left[\sqrt{2};2\right]}f\left(x\right)\) với \(f\left(x\right)=\dfrac{x^2+2x-10}{x+1}\)
Xét trên \(\left[\sqrt{2};2\right]\) ta có:
\(f\left(x\right)=\dfrac{3x^2+6x-30}{3\left(x+1\right)}=\dfrac{3x^2+8x-28-2\left(x+1\right)}{3\left(x+1\right)}=\dfrac{\left(3x+14\right)\left(x-2\right)}{3\left(x+1\right)}-\dfrac{2}{3}\le-\dfrac{2}{3}\)
\(\Rightarrow-2b\ge-\dfrac{2}{3}\Rightarrow b\le\dfrac{1}{3}\)
Vậy \(b_{max}=\dfrac{1}{3}\)
a, Đặt \(\sqrt[4]{a}=x;\sqrt[4]{b}=y.\)Bất đẳng thức ban đầu trở thành: \(\frac{2x^2y^2}{x^2+y^2}\le xy.\)
ta có : \(x^2+y^2\ge2xy\Rightarrow\frac{2x^2y^2}{x^2+y^2}\le\frac{2x^2y^2}{2xy}=xy.\)(đpcm )
dấu " = " xẩy ra khi x = y > 0
vậy bất đăng thức ban đầu đúng. dấu " = " xẩy ra khi a = b >0
Bài 1 :
Áp dụng bất đẳng thức Cauchy ta có :
\(\frac{\left(x-1\right)^2}{z}+\frac{z}{4}\ge2\sqrt{\frac{\left(x-1\right)^2}{z}\frac{z}{4}}=\left|x-1\right|=1-x\)
\(\frac{\left(y-1\right)^2}{x}+\frac{x}{4}\ge2\sqrt{\frac{\left(y-1\right)^2}{x}\frac{x}{4}}=\left|y-1\right|=1-y\)
\(\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge2\sqrt{\frac{\left(z-1\right)^2}{y}\frac{y}{4}}=\left|z-1\right|=1-z\)
\(\Rightarrow\frac{\left(x-1\right)^2}{z}+\frac{z}{4}+\frac{\left(y-1\right)^2}{x}+\frac{x}{4}+\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge1-x+1-y+1-z\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\ge3-\left(x+y+z\right)-\frac{x+y+z}{4}=3-2-\frac{2}{4}=\frac{1}{2}\)
Vậy GTNN của \(A=\frac{1}{2}\Leftrightarrow x=y=z=\frac{2}{3}\)
\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(\left|y\right|=\left|1.x+1.\sqrt{1-x^2}\right|\le\sqrt{\left(1^2+1^2\right)\left(x^2+1-x^2\right)}=\sqrt{2}\)