Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}-2b}{a-b}\)
\(=\dfrac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(c\ge\sqrt{ab}\Leftrightarrow\dfrac{c}{a}.\dfrac{c}{b}\ge1\)
BĐT cần chứng minh tương đương:
\(\dfrac{\left(c+a\right)^2}{c^2+a^2}\ge\dfrac{\left(c+b\right)^2}{c^2+b^2}\Leftrightarrow\dfrac{\left(\dfrac{c}{a}+1\right)^2}{\left(\dfrac{c}{a}\right)^2+1}\ge\dfrac{\left(\dfrac{c}{b}+1\right)^2}{\left(\dfrac{c}{b}\right)^2+1}\)
Đặt \(\left(\dfrac{c}{a};\dfrac{c}{b}\right)=\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}xy\ge1\\y>x\Rightarrow y-x>0\end{matrix}\right.\) (1)
BĐT cần c/m trở thành: \(\dfrac{\left(x+1\right)^2}{x^2+1}\ge\dfrac{\left(y+1\right)^2}{y^2+1}\Leftrightarrow\dfrac{x}{x^2+1}\ge\dfrac{y}{y^2+1}\)
\(\Leftrightarrow xy^2+x\ge x^2y+y\Leftrightarrow xy\left(y-x\right)-\left(y-x\right)\ge0\)
\(\Leftrightarrow\left(xy-1\right)\left(y-x\right)\ge0\) luôn đúng theo (1)
Vậy BĐT đã cho được c/m
Dấu "=" xảy ra khi \(xy=1\) hay \(c=\sqrt{ab}\)
Lời giải:
Biến đổi tương đương:
\(\sqrt{\frac{a+b}{2}}\geq \frac{\sqrt{a}+\sqrt{b}}{2}\)
\(\Leftrightarrow \frac{a+b}{2}\geq \frac{(\sqrt{a}+\sqrt{b})^2}{4}=\frac{a+b+2\sqrt{ab}}{4}\)
\(\Leftrightarrow \frac{a+b}{2}-\frac{a+b+2\sqrt{ab}}{4}\geq 0\)
\(\Leftrightarrow \frac{a+b-2\sqrt{ab}}{4}\geq 0\)
\(\Leftrightarrow \frac{(\sqrt{a}-\sqrt{b})^2}{4}\geq 0\) (luôn đúng)
Do đó ta có đpcm
Dấu "=" xảy ra khi $a=b$
a: \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\dfrac{1}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}=\sqrt{a}-\sqrt{b}\)
b: \(VT=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{2-\left(\sqrt{3}-1\right)}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{3-\sqrt{3}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)\left(\sqrt{3}-1\right)+2\left(\sqrt{2}-1\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\left(\sqrt{6}-\sqrt{2}+\sqrt{3}-1+\sqrt{6}+\sqrt{2}-\sqrt{3}-1\right)}{\sqrt{3}\cdot2}\)
\(=\dfrac{2\left(2\sqrt{6}-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}-2}{\sqrt{3}}\)
đúng với mọi a,b chứ nhỉ
nếu a, b <0 VT>=0 VP<0 => đúng
Bp
\(\Leftrightarrow2.\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow2a^2+2b^2-\left(a^2+2ab+b^2\right)\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) hiển nhiên đúng=> dpcm
a) \(\dfrac{a^2+3}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\ge2\sqrt{\sqrt{a^2+2}.\dfrac{1}{\sqrt{a^2+2}}}=2\)
Dấu = xảy ra khi \(\sqrt{a^2+2}=\dfrac{1}{\sqrt{a^2+2}}\Leftrightarrow a^2=-1\left(vn\right)\)
\(\Rightarrow\) Dấu "=" không xảy ra
Vậy \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)
b)Với x,y>0,ta cm bđt phụ sau:
\(x^3+y^3\ge xy\left(x+y\right)\) (1)
Thật vậy (1)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)
\(\Leftrightarrow\cdot\left(x+y\right)\left(x^2-2xy+y^2\right)\ge0\)
\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0\) (lđ)
Áp dụng (1) có:
\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}=\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}.\sqrt{b}}\ge\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" xra khi a=b
Vậy...
1) \(\left(a-b\right)^2\ge0\)
\(a^2-2ab+b^2\ge0\)
\(a^2+b^2+2ab\ge4ab\)
\(\left(a+b\right)^2\ge4ab\)
\(\dfrac{\left(a+b\right)^2}{4}\ge ab\)
\(\dfrac{a+b}{2}\ge\sqrt{ab}\)
Dấu ''='' xảy ra khi a=b
2) \(\left(\sqrt{2a}-\sqrt{2b}\right)^2\ge0\)
\(2a-4\sqrt{ab}+2b\ge0\)
\(4a+4b\ge2a+2b+4\sqrt{ab}\)
\(\dfrac{a+b}{2}\ge\dfrac{a+b+2\sqrt{ab}}{4}\)
\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
Dấu ''='' xảy ra khi a=b
\(\Leftrightarrow a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\)
\(\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
Vì (a-b)2\(\ge\)0 luôn đúng nên \(\sqrt{a^2+b^2}\ge\dfrac{a+b}{\sqrt{2}}\)