cho a > 0, b > 0. C/m :
\(\frac{3a^2+2ab+3b^2}{a+b}\ge2\sqrt{2\left(a^2+b^2\right)}\)
K
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12 tháng 11 2017
BĐT cần chứng minh tương đương
\(\dfrac{3a^2+2ab+3b^2}{a+b}-2\left(a+b\right)\ge2\sqrt{2\left(a^2+b^2\right)}-2\left(a+b\right)\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{a+b}\ge\dfrac{8\left(a^2+b^2\right)-4\left(a+b\right)^2}{2\sqrt{2\left(a^2+b^2\right)}+2\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{a+b}\ge\dfrac{2\left(a-b\right)^2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\)
\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{a+b}-\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\right)\ge0\)
ta phải chứng minh
\(\dfrac{1}{a+b}-\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\ge0\)
\(\Leftrightarrow\dfrac{1}{a+b}\ge\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\)
\(\Leftrightarrow\sqrt{2\left(a^2+b^2\right)}+a+b\ge2\left(a+b\right)\Leftrightarrow\sqrt{2\left(a^2+b^2\right)}\ge a+b\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
=> đpcm
LD
5
28 tháng 6 2020
ta có: \(\sqrt{4a\left(3a+b\right)}\le\frac{4a+3a+b}{2}=\frac{7a+b}{2}\)
=> \(\sqrt{a\left(3a+b\right)}\le\frac{7a+b}{4}\)
\(\sqrt{4b\left(3b+a\right)}\le\frac{7b+a}{4}\)
\(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{a+b}{\frac{7a+b}{4}+\frac{7b+a}{4}}=\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)
Dấu "=" xảy ra <=> a = b
28 tháng 6 2020
Sửa đề: CM: \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{1}{2}\)
Ta có \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\left(1\right)\)
Áp dụng bất đẳng thức Cô-si cho các só dương ta được
\(\hept{\begin{cases}\sqrt{4a\left(3a+b\right)}\le\frac{4a+\left(3a+b\right)}{2}=\frac{7a+b}{2}\left(2\right)\\\sqrt{4b\left(3b+a\right)}\le\frac{4b+\left(3b+a\right)}{2}=\frac{7b+a}{2}\left(3\right)\end{cases}}\)
Từ (2) và (3) \(\Rightarrow\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}\le4a+4b\left(4\right)\)
Từ (1) và (4) => \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{2\left(a+b\right)}{4a+4b}=\frac{1}{2}\)
Dấu "=" xảy ra <=> a=b
QT
25 tháng 6 2021
+) 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 :
\(VT\left(1\right)\ge\frac{a+b}{\frac{7a+b}{4}+\frac{7b+a}{4}}=\frac{1}{2}\left(đpcm\right)\)
25 tháng 6 2021
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
25 tháng 2 2017
Ta có: \(\sqrt{3a^2+8b^2+14ab}=\sqrt{\left(3a+2b\right)\left(a+4b\right)}\le2a+3b\)
Khi đó \(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\ge\frac{a^2}{2a+3b}\), tương tự cho ta cũng có:
\(\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}\ge\frac{b^2}{2b+3c};\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\ge\frac{c^2}{2c+3a}\)
Cộng theo vế ta có: \(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\)
\(\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)
KK
25 tháng 2 2017
\(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{3a^2+12ab+8b^2+2ab}}+\frac{b^2}{\sqrt{3b^2+12bc+8c^2+2bc}}+\frac{c^2}{\sqrt{3c^2+12ca+8a^2+2ca}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{3a\left(a+4b\right)+2b\left(4b+a\right)}}+\frac{b^2}{\sqrt{3b\left(b+4c\right)+2c\left(4c+b\right)}}+\frac{c^2}{\sqrt{3c\left(c+4a\right)+2a\left(4a+c\right)}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{\left(a+4b\right)\left(3a+2b\right)}}+\frac{b^2}{\sqrt{\left(b+4c\right)\left(3b+2c\right)}}+\frac{c^2}{\sqrt{\left(c+4a\right)\left(3c+2a\right)}}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\sqrt{\left(a+4b\right)\left(3a+2b\right)}\le\frac{4a+6b}{2}\\\sqrt{\left(b+4c\right)\left(3b+2c\right)}\le\frac{4b+6c}{2}\\\sqrt{\left(c+4a\right)\left(3c+2a\right)}\le\frac{4c+6a}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a^2}{\sqrt{\left(a+4b\right)\left(3a+2b\right)}}\ge\frac{2a^2}{4a+6b}\\\frac{b^2}{\sqrt{\left(b+4c\right)\left(3b+2c\right)}}\ge\frac{2b^2}{4b+6c}\\\frac{c^2}{\sqrt{\left(c+4a\right)\left(3c+2a\right)}}\ge\frac{2c^2}{4c+6a}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{2a^2}{4a+6b}+\frac{2b^2}{4b+6c}+\frac{2c^2}{4c+6a}\)
Chứng minh rằng \(\frac{2a^2}{4a+6b}+\frac{2b^2}{4b+6c}+\frac{2c^2}{4c+6a}\ge\frac{1}{5}\left(a+b+c\right)\)
\(\Leftrightarrow2\left(\frac{a^2}{4a+6b}+\frac{b^2}{4b+6c}+\frac{c^2}{4c+6a}\right)\ge\frac{1}{5}\left(a+b+c\right)\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{a^2}{4a+6b}+\frac{b^2}{4b+6c}+\frac{c^2}{4c+6a}\ge\frac{\left(a+b+c\right)^2}{10\left(a+b+c\right)}\)
\(\Rightarrow2\left(\frac{a^2}{4a+6b}+\frac{b^2}{4b+6c}+\frac{c^2}{4c+6a}\right)\ge\frac{2\left(a+b+c\right)^2}{10\left(a+b+c\right)}=\frac{a+b+c}{5}\)
\(\Rightarrow2\left(\frac{a^2}{4a+6b}+\frac{b^2}{4b+6c}+\frac{c^2}{4c+6a}\right)\ge\frac{1}{5}\left(a+b+c\right)\)
Vậy \(\frac{2a^2}{4a+6b}+\frac{2b^2}{4b+6c}+\frac{2c^2}{4c+6a}\ge\frac{1}{5}\left(a+b+c\right)\)
Mà \(VT\ge\frac{2a^2}{4a+6b}+\frac{2b^2}{4b+6c}+\frac{2c^2}{4c+6a}\)
\(\Rightarrow VT\ge\frac{1}{5}\left(a+b+c\right)\)
\(\Leftrightarrow\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\ge\frac{1}{5}\left(a+b+c\right)\)
( đpcm )
8 tháng 1 2020
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm