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\(A=\frac{a^2+3ab+b^2}{\sqrt{ab}\left(a+b\right)}=\frac{\left(a^2+2ab+b^2\right)+ab}{\sqrt{ab}\left(a+b\right)}=\frac{\left(a+b\right)^2+ab}{\sqrt{ab}\left(a+b\right)}\)
\(=\frac{\left(a+b\right)^2}{\sqrt{ab}\left(a+b\right)}+\frac{ab}{\sqrt{ab}\left(a+b\right)}=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)
Áp dụng bđt AM - GM ta có : \(A\ge2\sqrt{\frac{a+b}{\sqrt{ab}}.\frac{\sqrt{ab}}{a+b}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow a+b=\sqrt{ab}\)
làm tiếp đoạn của Đinh Đức Hùng
\(\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}=\frac{a+b}{\sqrt{ab}}+\frac{4\sqrt{ab}}{a+b}-\frac{3\sqrt{ab}}{a+b}\ge4-\frac{\frac{3}{2}\left(a+b\right)}{a+b}=4-\frac{3}{2}=\frac{5}{2}\)
\(ab\cdot\sqrt{\dfrac{a}{3b}}-a^2\sqrt{\dfrac{3b}{a}}\)
\(=a\sqrt{ab}-a^2\cdot\dfrac{\sqrt{3b}}{\sqrt{a}}\)
\(=a\sqrt{ab}-a\sqrt{a}\cdot\sqrt{3b}\)
\(=a\sqrt{ab}\left(1-\sqrt{3}\right)\)
\(\Leftrightarrow m=\dfrac{a\sqrt{ab}\left(1-\sqrt{3}\right)}{\sqrt{3ab}}=\dfrac{a\left(\sqrt{3}-3\right)}{3}\)
Câu 1 : áp dụng BĐT SVAC ta có \(A\ge\frac{(a+b+c)^2}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}}=\frac{1.\sqrt{2a+2b+2c}}{\sqrt{2.}(\sqrt{b+c}+\sqrt{a+b}+\sqrt{a+c})}\)
mặt khác lại có \(\frac{\sqrt{2a+2b+2c}}{\sqrt{2}.(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})}\ge\frac{\sqrt{(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})^2}}{\sqrt{2}.\sqrt{3}.(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})}=\frac{1}{\sqrt{6}}\)theo bđt svac
\(\Rightarrow A\ge\frac{1}{\sqrt{6}}\)dấu bằng xảy ra tại a=b=c=\(\frac{1}{3}\)
Sửa đề: Tìm Max của \(\frac{1}{\sqrt{a^2+1}}+\frac{2}{\sqrt{b^2+4}}+\frac{3}{\sqrt{c^2+9}}\) biết a,b,c>0 và 6a+3b+2c=abc
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{\sqrt{a^3}-3a\sqrt{b}+3\sqrt{a}.b-\sqrt{b^3}+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{3\sqrt{a^3}-3a\sqrt{b}+3b\sqrt{a}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)
\(=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)
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 <=> \(a=b=c=\sqrt{3}\)
