Cho \(a>0;b>0\) .Chứng minh:
\(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge2\)
Với mọi a, b>0 chứng minh \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{1}{2}\)
\(\frac{\left(a+b\right).2}{\sqrt{a.4.\left(3a+b\right)}+\sqrt{b.4.\left(3b+a\right)}}\)\(\ge\)\(\frac{2.\left(a+b\right)}{\frac{4a+3a+b}{2}+\frac{4b+3b+a}{2}}\)\(=\frac{4\left(a+b\right)}{8\left(a+b\right)}=\frac{1}{2}\)
Dấu "=" xảy ra khi và chỉ khi a=b
cho a > 0, b > 0. C/m :
\(\frac{3a^2+2ab+3b^2}{a+b}\ge2\sqrt{2\left(a^2+b^2\right)}\)
Chứng minh:
\(\frac{a+b}{\sqrt{a\left(a+3b\right)}+\sqrt{b\left(b+3a\right)}}\)\(\ge\)\(\frac{1}{2}\) (a,b>0)
Ta có:
\(\frac{a+b}{\sqrt{a\left(a+3b\right)}+\sqrt{b\left(b+3a\right)}}=\frac{2\left(a+b\right)}{\sqrt{4a\left(a+3b\right)}+\sqrt{4b\left(b+3a\right)}}\) (nhân 2 vào cả tử và mẫu)
\(\ge\frac{2\left(a+b\right)}{\frac{4a+a+3b}{2}+\frac{4b+b+3a}{2}}=\frac{4\left(a+b\right)}{8\left(a+b\right)}=\frac{1}{2}^{\left(đpcm\right)}\) (áp dụng BĐT Cô si vào cái mẫu)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}4a=a+3b\\4b=b+3a\end{matrix}\right.\Leftrightarrow a=b\)
Áp dụng BĐT Côsi ta có:
\( \sqrt {4a\left( {3a + b} \right)} \le \dfrac{{4a + 3a + b}}{2} = \dfrac{{7a + b}}{2}\\ \Rightarrow \sqrt {a\left( {3a + b} \right)} \le \dfrac{{7a + b}}{4}\\ \sqrt {4b\left( {3b + a} \right)} \le \dfrac{{4b + 3b + a}}{2} = \dfrac{{7b + a}}{2}\\ \Rightarrow \sqrt {b\left( {3b + a} \right)} \le \dfrac{{7b + a}}{4}\\ \Rightarrow \sqrt {a\left( {3a + b} \right)} + \sqrt {b\left( {3b + a} \right)} \le \dfrac{{7b + a + 7a + b}}{4} = 2\left( {a + b} \right)\\ \Rightarrow \dfrac{{a + b}}{{\sqrt {a\left( {3a + b} \right)} + \sqrt {b\left( {3b + a} \right)} }} \ge \dfrac{1}{2} \)
Dấu "=" xảy ra\(\left\{{}\begin{matrix}4a=3a+b\\4b=3b+a\end{matrix}\right.\Leftrightarrow a=b\)
Cho a,b > 0 . CMR : \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{1}{2}\left(1\right)\)
+) Ta có \(\sqrt{4a\left(3a+b\right)}\le\frac{4a+\left(3a+b\right)}{2}=\frac{7a+b}{2}\)
\(\Rightarrow\sqrt{a\left(3a+b\right)}\le\frac{7a+b}{4}\left(2\right)\)
+) Tương tự ta lại có :
\(\sqrt{b\left(3b+a\right)}\le\frac{7b+a}{4}\left(3\right)\)
+) Từ (2) và (3) ta có :
\(VT\left(1\right)\ge\frac{a+b}{\frac{7a+b}{4}+\frac{7b+a}{4}}=\frac{1}{2}\left(đpcm\right)\)
Ta có: \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\)
\(=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\ge\frac{2\left(a+b\right)}{\frac{1}{2}\left(4a+3a+b\right)+\frac{1}{2}\left(4b+3b+a\right)}\) (Cauchy)
\(=\frac{2\left(a+b\right)}{4\left(a+b\right)}=\frac{1}{2}\)
Dấu "=" xảy ra khi: a = b
Với a, b > 0 và biểu thức \(A=\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\) . Hãy chứng minh \(A\ge\frac{1}{2}\)
Với a , b > 0 . Ta có : \(\left(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\right)^2\le\left(\sqrt{a}^2+\sqrt{b}^2\right)\left(\sqrt{3a+b}^2+\sqrt{3b+a}^2\right)= \left(a+b\right).4\left(a+b\right)\)
\(\Rightarrow\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\le2\left(a+b\right)\) ( vì a , b > 0 )
\(\Rightarrow A\ge\frac{1}{2}\left(đpcm\right)\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{3a+b}{a}=\frac{3b+a}{b}\Leftrightarrow a=b\)
CMR \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\)\(\ge2\)
Bài 1: Cho a, b cùng dấu. Chứng minh rằng: \(\left(\frac{a^2+b^2}{2}\right)^3\le\left(\frac{a^3+b^3}{2}\right)^2\)
Bài 2: Cho \(a^2+b^2\ne0\). Chứng minh rằng: \(\frac{2ab}{a^2+4b^2}+\frac{b^2}{3a^2+2b^2}\le\frac{3}{5}\)
Bài 3: Cho a, b > 0. Chứng minh rằng: \(\frac{a}{b^2}+\frac{b}{a^2}+\frac{16}{a+b}\ge5\left(\frac{1}{a}+\frac{1}{b}\right)\)
Bài 4: Cho a, b>0. Chứng minh rằng: \(\frac{3a^2+2ab+3b^2}{a+b}\ge2\sqrt{2\left(a^2+b^2\right)}\)
Bài 1: Cho a,b>0. Chứng minh \(\sqrt[3]{\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)}< \sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\)
Bài 2: Cho a,b>0. Chứng minh \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\ge\frac{2\sqrt{2}}{\sqrt{a+b}}\)
Bài 3: Cho a,b,c>0. Chứng minh \(\frac{5b^3-a^3}{ab+3b^2}+\frac{5c^3-b^3}{bc+3c^2}+\frac{5a^3-c^3}{ca+3a^2}\le a+b+c\)
Cho a, b>0. Chứng minh rằng:
a) \(\dfrac{3a^2+2ab+3b^2}{a+b}\ge2\sqrt{2\left(a^2+b^2\right)}\)
b) \(\dfrac{2ab}{a+b}+\sqrt{\dfrac{a^2+b^2}{2}}\ge\sqrt{ab}+\dfrac{a+b}{2}\)
c) \(\dfrac{1}{\left(1+a\right)^2}+\dfrac{1}{\left(1+b\right)^2}\ge\dfrac{1}{1+ab}\)