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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
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
+) 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
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{\sqrt{a^3}-3a\sqrt{b}+3\sqrt{a}.b-\sqrt{b^3}+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{3\sqrt{a^3}-3a\sqrt{b}+3b\sqrt{a}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)
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: \(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\)
\(\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)
\(\frac{a^2}{\sqrt{3a^2+8b^2+14ab}}+\frac{b^2}{\sqrt{3b^2+8c^2+14bc}}+\frac{c^2}{\sqrt{3c^2+8a^2+14ca}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{3a^2+12ab+8b^2+2ab}}+\frac{b^2}{\sqrt{3b^2+12bc+8c^2+2bc}}+\frac{c^2}{\sqrt{3c^2+12ca+8a^2+2ca}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{3a\left(a+4b\right)+2b\left(4b+a\right)}}+\frac{b^2}{\sqrt{3b\left(b+4c\right)+2c\left(4c+b\right)}}+\frac{c^2}{\sqrt{3c\left(c+4a\right)+2a\left(4a+c\right)}}\)
\(\Leftrightarrow\frac{a^2}{\sqrt{\left(a+4b\right)\left(3a+2b\right)}}+\frac{b^2}{\sqrt{\left(b+4c\right)\left(3b+2c\right)}}+\frac{c^2}{\sqrt{\left(c+4a\right)\left(3c+2a\right)}}\)
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 )