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Ta có:
\(21b+\frac{3}{a}=\frac{3}{a}+\frac{a}{3}+\frac{62a}{3}\ge2\sqrt{\frac{3}{a}.\frac{a}{3}}+\frac{62.3}{3}=2+62=64\left(a\ge3\right)\left(1\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{3}{a}=\frac{a}{3}\)và \(a=3\Leftrightarrow a=3\)
\(\frac{21}{b}+3b=\frac{21}{b}+\frac{7b}{3}+\frac{2b}{3}\ge2\sqrt{\frac{21}{b}.\frac{7b}{3}}+\frac{2.3}{3}=14+2=16\left(b\ge3\right)\left(2\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{21}{b}=\frac{7b}{3}\)và \(b=3\Leftrightarrow b=3\)
Từ (1) và (2) suy ra điều cần chứng minh.
Dấu "=" xảy ra \(\Leftrightarrow a=b=3\)
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\(BDT\Leftrightarrow\dfrac{\left(x^2-y^2\right)^2}{x^2y^2}\ge\dfrac{3\left(x-y\right)^2}{xy}\)
\(\Leftrightarrow\dfrac{\left[\left(x-y\right)\left(x+y\right)\right]^2}{x^2y^2}-\dfrac{3\left(x-y\right)^2}{xy}\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{\left(x+y\right)^2}{x^2y^2}-\dfrac{3}{xy}\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{\left(x+y\right)^2-3xy}{x^2y^2}\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(\dfrac{x^2+y^2-xy}{x^2y^2}\right)\ge0\) (luôn đúng)
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Áp dụng bất đẳng thức AM-GM ta có :
\(x+\frac{1}{2x}\ge2\sqrt{x\cdot\frac{1}{2x}}=2\sqrt{\frac{1}{2}}\)
\(y+\frac{2}{y}\ge2\sqrt{y\cdot\frac{2}{y}}=2\sqrt{2}\)
=> \(x+\frac{1}{2x}+y+\frac{2}{y}\ge2\sqrt{\frac{1}{2}}+2\sqrt{2}=3\sqrt{2}\left(đpcm\right)\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x=\sqrt{\frac{1}{2}}\\y=\sqrt{2}\end{cases}}\)
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\(\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)
Tương tự \(\dfrac{\sqrt{1+y^3+z^3}}{yz}\ge\dfrac{\sqrt{3}}{\sqrt{yz}};\dfrac{\sqrt{1+x^3+z^3}}{xz}\ge\dfrac{\sqrt{3}}{\sqrt{xz}}\)
\(\Rightarrow VT\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3}.\dfrac{3}{\sqrt[3]{xyz}}=3\sqrt{3}\)
Dấu "=" xảy ra khi x=y=z=1
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Note \(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+2\)
Nên ta sẽ đặt \(\dfrac{x}{y}+\dfrac{y}{x}=t\ge2\). Khi đó
\(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2+2\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(t^2+2\ge3t\Leftrightarrow\left(t-2\right)\left(t-1\right)\ge0\)
BĐT cuối đúng vì \(t\ge 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức AM-GM ta có :
\(x+\frac{1}{x-1}=\left[\left(x-1\right)+\frac{1}{x-1}\right]+1\ge2\sqrt{\left(x-1\right)\cdot\frac{1}{x-1}}+1=2+1=3\left(đpcm\right)\)
Đẳng thức xảy ra <=> x = 2