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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 <=> \(a=b=c=\sqrt{3}\)
\(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{bc+a\left(a+b+c\right)}}=a\sqrt{\frac{1}{a+b}.\frac{1}{c+a}}\le\frac{\frac{a}{a+b}+\frac{a}{c+a}}{2}\)
Tương tự 2 cái còn lại cộng lại ta đc \(VT\le\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
Cach khac
Dat \(P=\frac{a}{\sqrt{bc\left(1+a^2\right)}}+\frac{b}{\sqrt{ca\left(1+b^2\right)}}+\frac{c}{\sqrt{ab\left(1+c^2\right)}}\)
Ta co:
\(a+b+c=abc\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Dat \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)
\(\Rightarrow xy+yz+zx=1\)
\(\Rightarrow P=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
Ta lai co:
\(\sqrt{\frac{yz}{1+x^2}}=\sqrt{\frac{yz}{xy+yz+zx+x^2}}=\sqrt{\frac{yz}{\left(x+y\right)\left(z+x\right)}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{z+x}\right)\)
Tuong tu:
\(\sqrt{\frac{zx}{1+y^2}}\le\frac{1}{2}\left(\frac{z}{y+z}+\frac{x}{x+y}\right)\)
\(\sqrt{\frac{xy}{1+z^2}}\le\frac{1}{2}\left(\frac{x}{z+x}+\frac{y}{y+z}\right)\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{2}\)
Dau '=' xay ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
\(\Rightarrow a=b=c=\sqrt{3}\)
Vay \(P_{min}=\frac{3}{2}\)khi \(a=b=c=\sqrt{3}\)
\(a^2-ab+b^2=\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2\ge\frac{1}{4}\left(a+b\right)^2\)
\(\frac{1}{\sqrt{a^2-ab+b^2}}\le\frac{1}{\sqrt{\frac{1}{4}\left(a+b\right)^2}}=\frac{2}{a+b}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Tương tự với 2 số còn lại, cộng theo vế ta được kết quả cần tìm.
\(\sqrt{c+ab}\) =\(\sqrt{c\left(a+b+c\right)+ab}=\sqrt{c^2+ac+cb+ab}=\sqrt{\left(c+a\right)\left(c+b\right)}\)
\(\frac{ab}{\sqrt{c+ab}}\le\frac{ab}{2}\left(\frac{1}{c+a}+\frac{1}{b+c}\right)\)
ttu \(\frac{bc}{\sqrt{a+bc}}\le\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{a+c}\right);\frac{ac}{\sqrt{b+ca}}\le\frac{1}{2}\left(\frac{1}{b+a}+\frac{1}{a+c}\right)\)
\(\Rightarrow P\le\frac{bc+ac}{2\left(a+b\right)}+\frac{ac+ab}{2\left(a+b\right)}+\frac{bc+ab}{2\left(c+b\right)}=\frac{1}{2}\left(a+b+c\right)=\frac{1}{2}\)
dau = xay ra khi a=b=c=1/3
\(P=\frac{\sqrt{a-1}}{a}+\frac{\sqrt{b-4}}{b}+\frac{\sqrt{c-9}}{c}\)
Vì \(a=\left(a-1\right)+1\ge2\sqrt{\left(a-1\right).1}=2\sqrt{a-1}\)
\(b=\left(b-4\right)+4\ge2\sqrt{\left(b-4\right).4}=4\sqrt{b-4}\)
\(c=\left(c-9\right)+9\ge2\sqrt{\left(c-9\right).9}=6\sqrt{c-9}\)
=>\(P\le\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}\)
P max = 11/12 khi a=2; b=8; c =18