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\(A=\dfrac{a}{\sqrt{a-1}}=\dfrac{a-1+1}{\sqrt{a-1}}=\sqrt{a-1}+\dfrac{1}{\sqrt{a-1}}\ge2\sqrt{\dfrac{\sqrt{a-1}}{\sqrt{a-1}}}=2\)
\(A_{min}=2\) khi \(a-1=1\Leftrightarrow a=2\)
\(\frac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow\frac{a+b}{2}-\sqrt{ab}\ge0\)
\(\Leftrightarrow\frac{a+b-2\sqrt{ab}}{2}\ge0\)
\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}\ge0\) (luôn đúng)
Vậy \(\frac{a+b}{2}\ge\sqrt{ab}\) (1)
\(\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\)
\(\Leftrightarrow\sqrt{ab}\ge\frac{2ab}{a+b}\)
\(\Leftrightarrow\sqrt{ab}\ge\frac{2\sqrt{ab}^2}{a+b}\)
\(\Leftrightarrow\frac{2\sqrt{ab}}{a+b}\le1\)
\(\Leftrightarrow\frac{2\sqrt{ab}}{a+b}-1\le0\)
\(\Leftrightarrow\frac{2\sqrt{ab}-a-b}{a+b}\le0\)
\(\Leftrightarrow\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{a+b}\le0\) (luôn đúng)
Vậy \(\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a+b}{2}\ge\sqrt{ab}\ge\frac{2}{\frac{1}{a}+\frac{1}{b}}\) (đpcm)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{2}\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(\ge\frac{1}{2}\frac{4}{a+b}+\frac{1}{2}\frac{4}{b+c}+\frac{1}{2}\frac{4}{c+a}\)
\(=\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\)
Dấu "=" xảy ra <=> a = b = c
a) Để \(a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\left(đpcm\right)\)
b) Để \(a\left(a+2\right)< \left(a+1\right)^2\)
\(\Leftrightarrow a^2+2a< a^2+2a+1\)
\(\Leftrightarrow a^2+2a-a^2-2a< a^2+2a+1-a^2-2a\)
\(\Leftrightarrow0< 1\left(đpcm\right)\)
c) Cách 1 : Để \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b}\ge0\)
\(\Leftrightarrow\frac{b+a}{ab}-\frac{4}{a+b}\ge0\)
\(\Leftrightarrow\frac{a^2+2ab+b^2-4ab}{ab\left(a+b\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\left(đpcm\right)\)
Cách 2 : Vì a > 0, b > 0
Áp dụng bất đẳng thức Cô-si dạng phân thức ta có :
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\left(đpcm\right)\)
Bài làm:
Ta có: \(A=x+\frac{1}{x^2}=\left(\frac{1}{x^2}+\frac{x}{8}+\frac{x}{8}\right)+\frac{3}{4}x\ge3\sqrt[3]{\frac{1}{x^2}.\frac{x}{8}.\frac{x}{8}}+\frac{3}{4}.2\)
\(=3.\frac{1}{4}+\frac{3}{2}=\frac{3}{4}+\frac{3}{2}=\frac{9}{4}\)
Dấu "=" xảy ra khi: \(\frac{1}{x^2}=\frac{x}{8}\Leftrightarrow x^3=8\Leftrightarrow x=2\)
Vậy \(Min\left(A\right)=\frac{9}{4}\)khi \(x=2\)
Học tốt!!!!
Đặt \(x=\frac{a}{b}+\frac{b}{a}\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{a^2}=x^2-2\)
Xét mẫu thức : \(\frac{a^2}{b^2}+\frac{b^2}{a^2}-\left(\frac{a}{b}+\frac{b}{a}\right)=x^2-x-2=\left(x+1\right)\left(x-2\right)\)
Thay \(x=\frac{a}{b}+\frac{b}{a}\) được mẫu thức : \(\left(\frac{a}{b}+\frac{b}{a}+1\right)\left(\frac{a}{b}+\frac{b}{a}-2\right)=\left(\frac{a}{b}+\frac{b}{a}+1\right).\frac{\left(a-b\right)^2}{ab}\)
Ta có : \(P=\frac{\left(\frac{a}{b}+\frac{b}{a}+1\right)\left(\frac{1}{a}-\frac{1}{b}\right)^2}{\frac{a^2}{b^2}+\frac{b^2}{a^2}-\left(\frac{a}{b}+\frac{b}{a}\right)}=\frac{\left(\frac{a}{b}+\frac{b}{a}+1\right).\frac{\left(a-b\right)^2}{a^2b^2}}{\left(\frac{a}{b}+\frac{b}{a}+1\right).\frac{\left(a-b\right)^2}{ab}}\)
\(=\frac{\left(a-b\right)^2}{a^2b^2}.\frac{ab}{\left(a-b\right)^2}=\frac{1}{ab}\) (đpcm)
b) Áp dụng bđt Cauchy :
\(1=4a+b+\sqrt{ab}\ge2\sqrt{4a.b}+\sqrt{ab}\)
\(\Rightarrow5\sqrt{ab}\le1\Rightarrow ab\le\frac{1}{25}\)
\(\Rightarrow P=\frac{1}{ab}\ge25\) . Dấu "=" xảy ra khi \(\begin{cases}4a+b+\sqrt{ab}=1\\4a=b\end{cases}\)
\(\Leftrightarrow\begin{cases}a=\frac{1}{10}\\b=\frac{2}{5}\end{cases}\)
Vậy P đạt giá trị nhỏ nhất bằng 25 tại \(\left(a;b\right)=\left(\frac{1}{10};\frac{2}{5}\right)\)
pn ơi , bđt cauchy : \(a+b\ge2\sqrt{ab}\)
s lại là \(2\sqrt{4a.b}+\sqrt{ab}\)
CM theo bdt co-si
Áp dụng bdt Co - si cho cặp số dương a2/c và c
Ta có: \(\frac{a^2}{c}+c\ge2\sqrt{\frac{a^2}{c}.c}=2a\)(1)
CMTT: \(\frac{b^2}{a}+a\ge2b\)(2)
\(\frac{c^2}{b}+b\ge2c\)(3)
Từ (1); (2) và (3) cộng vế theo vế, ta có:
\(\frac{a^2}{c}+c+\frac{b^2}{a}+a+\frac{c^2}{b}+b\ge2a+2b+2c\)
<=> \(\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\ge2a+2b+2c-a-b-c=a+b+c\)(Đpcm)
\(\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra <=> a = b = c