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Ta có:\(\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{a\left(a+b\right)+c\left(a+b\right)}}\)
\(=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\) (Áp dụng BĐT AM-GM)
Tương tự với hai BĐT còn lại và cộng theo vế ta thu được đpcm.
Tham khảo
Câu hỏi của Châu Trần - Toán lớp 9 - Học toán với OnlineMath
Đặt vế trái là P và \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=4\)
Ta cần chứng minh: \(P=\frac{1}{xy+2yz+zx}+\frac{1}{xy+yz+2zx}+\frac{1}{2xy+yz+zx}\le\frac{1}{xyz}\)
\(P=\frac{1}{xy+yz+yz+zx}+\frac{1}{xy+yz+zx+zx}+\frac{1}{xy+xy+yz+zx}\)
\(P\le\frac{1}{16}\left(\frac{1}{xy}+\frac{2}{yz}+\frac{1}{zx}+\frac{1}{xy}+\frac{1}{yz}+\frac{2}{zx}+\frac{2}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(P\le\frac{1}{4}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{4}\left(\frac{x+y+z}{xyz}\right)=\frac{1}{4}.\frac{4}{xyz}=\frac{1}{xyz}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=\frac{4}{3}\) hay \(a=b=c=\frac{16}{9}\)
1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)
\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)
\(VT=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{a}{a+c}+\frac{b}{b+c}}{2}\)
Tượng tự ta có \(\hept{\begin{cases}\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\le\frac{\frac{c}{b+c}+\frac{a}{a+b}}{2}\end{cases}}\)
\(\Rightarrow VT\le\frac{\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\left(\frac{c}{a+c}+\frac{a}{c+a}\right)+\left(\frac{c}{b+c}+\frac{b}{c+b}\right)}{2}\)
\(\Rightarrow VT\le\frac{\frac{a+b}{a+b}+\frac{c+a}{c+a}+\frac{b+c}{b+c}}{2}=\frac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)
Ta có:
\(\frac{a}{\sqrt{1+a^2}}=\frac{a}{\sqrt{a^2+ab+bc+ac}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Sau đó Cauchy....
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Do \(a+b+c=1\) nên :
\(VT=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\frac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\frac{ca}{b\left(a+b+c\right)+ac}}\)
\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)
Áp dụng BĐT AM - GM :
\(\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right)\)
\(\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)
Cộng theo vế :
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt !!!
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:
\(VT=\Sigma_{cyc}\sqrt{\frac{yz}{\left(xy+yz\right)+\left(xz+yz\right)}}\le\Sigma_{cyc}\sqrt{\frac{yz}{4}\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)}\)
\(=\frac{1}{2}\Sigma_{cyc}\sqrt{1\left(\frac{yz}{xy+yz}+\frac{yz}{xz+yz}\right)}\le\frac{1}{4}\Sigma_{cyc}\left(1+\frac{yz}{xy+yz}+\frac{yz}{xz+yz}\right)=\frac{3}{2}\)
\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)
Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)
