Ta có: \(3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2=\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\)
\(\Rightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\) nên với \(x,y,z>0\) ta có:
\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\) áp dụng ta có:
\(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}}\le\sqrt{3\left(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\right)}\)
Với: \(x,y>0\) ta có: \(x+y\ge2\sqrt{xy}\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng ta được:
\(\frac{1}{ab+a+2}=\frac{1}{ab+1+a+1}=\frac{1}{ab+abc+a+1}=\frac{1}{ab\left(c+1\right)+\left(a+1\right)}\)
\(\le\frac{1}{4}\left(\frac{1}{ab\left(c+1\right)}+\frac{1}{a+1}\right)=\frac{1}{4}\left(\frac{abc}{ab\left(c+1\right)}+\frac{1}{a+1}\right)=\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)
Vậy ta có: \(\frac{1}{ab+a+2}\le\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)
Tương tự như trên ta có: \(\frac{1}{bc+b+2}\le\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{b+1}\right)\) và \(\frac{1}{ca+c+2}\le\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{c+1}\right)\) nên:
\(\Rightarrow\sqrt{3\left(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\right)}\)
\(\le\sqrt{3.\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}+\frac{a}{a+1}+\frac{1}{b+1}+\frac{b}{b+1}+\frac{1}{c+1}\right)}=\frac{3}{2}\)
Vậy \(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}}\le\frac{3}{2}\left(đpcm\right)\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Đặt \(\left(a;b;c\right)=\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\). BĐT quy về:\(\Sigma_{cyc}\frac{\sqrt{yz}}{\sqrt{xy+xz+2yz}}\le\frac{3}{2}\)
Áp dụng liên hoàn BĐT Cô si:
\(VT=\Sigma_{cyc}\sqrt{\frac{yz}{\left(xy+yz\right)+\left(xz+yz\right)}}\le\Sigma_{cyc}\sqrt{\frac{yz}{4}\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)}\)
\(=\frac{1}{2}\Sigma_{cyc}\sqrt{1\left(\frac{yz}{xy+yz}+\frac{yz}{xz+yz}\right)}\le\frac{1}{4}\Sigma_{cyc}\left(1+\frac{yz}{xy+yz}+\frac{yz}{xz+yz}\right)=\frac{3}{2}\)
Áp dụng bất đẳng thức Cô - si ta có:
\(\frac{1}{\sqrt{ab+a+2}}\le\left(\frac{1}{4}+\frac{1}{ab+a+2}\right)\)
Tương tự:
=> \(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ac+c+2}}\)
\(\le\frac{3}{4}+\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\)(1)
Áp dụng: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) với x, y >0
Ta có: \(\frac{1}{ab+a+2}=\frac{1}{\frac{ab}{abc}+a+2}\le\frac{1}{4}.\left(\frac{1}{\frac{1}{c}+1}+\frac{1}{a+1}\right)\)vì abc =1
\(=\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)
Tương tự
=> \(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\)
\(\le\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)+\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{b+1}\right)+\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{c+1}\right)\)
\(=\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{c+1}+\frac{1}{a+1}+\frac{a}{a+1}+\frac{1}{b+1}+\frac{b}{b+1}\right)\)
\(=\frac{1}{4}\left(1+1+1\right)=\frac{3}{4}\)(2)
Từ (1); (2)
=> \(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ac+c+2}}\le\frac{3}{4}+\frac{3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1
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Từ \(abc=1\)suy ra tồn tại \(x,y,z>0\)sao cho \(\left(a,b,c\right)=\left(\frac{x}{y},\frac{y}{z},\frac{z}{x}\right)\)
Bài toán chuyển về CMR :
\(A=\sqrt{\frac{yz}{xy+xz+2yz}}+\sqrt{\frac{xz}{xy+yz+2xz}}+\sqrt{\frac{xy}{2xy+yz+xz}}\le\frac{3}{4}\)
Áp dụng bất đẳng thức AM - GM :
\(\sqrt{\frac{yz}{xy+xz+2yz}}\le\frac{yz}{xy+xz+2yz}+\frac{1}{4}\)
Thiết lập tương tự ..
\(\Rightarrow A\le\frac{xy}{2xy+yz+xz}+\frac{yz}{xy+2yz+xz}+\frac{xz}{xy+yz+2xz}+\frac{3}{4}\left(1\right)\)
Áp dụng BĐT Cauchy Schwarz :
\(\frac{1}{\frac{xy+yz+xz}{3}}+\frac{1}{\frac{xy+yz+xz}{3}}+\frac{1}{\frac{xy+yz+xz}{3}}+\frac{1}{xy}\ge\frac{16}{2xy+yz+xz}\Rightarrow\frac{9}{xy+yz+xz}+1\)
\(\ge\frac{16xy}{2xy+yz+xz}\)
Thiết lập tương tự với các phân thức còn lại và cộng theo vế :
\(\Rightarrow\frac{xy}{2xy+yz+xz}+\frac{yz}{xy+2yz+xz}+\frac{xz}{xy+yz+2xz}\le\frac{12}{16}=\frac{3}{4}\left(2\right)\)
Từ 1 và 2 \(\Rightarrow A\le\frac{3}{2}\left(đpcm\right)\)
Dấu " = " xảy ra khi x=y=z hay a=b=c=1
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