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\(x+y+z=1\\ \Rightarrow\left\{{}\begin{matrix}x=1-y-z\\y=1-x-z\\z=1-x-y\end{matrix}\right.\)
\(S=\dfrac{\left(xy+z\right)\left(yz+x\right)\left(zx+y\right)}{\left(1-x\right)^2\left(1-y\right)^2\left(1-z\right)^2}\)
\(\Rightarrow S=\dfrac{\left(xy+1-x-y\right)\left(yz+1-y-z\right)\left(zx+1-x-z\right)}{\left(x+y+z-x\right)^2\left(x+y+z-y\right)^2\left(x+y+z-z\right)^2}\)
\(\Rightarrow S=\dfrac{\left[\left(xy-x\right)-\left(y-1\right)\right]\left[\left(yz-y\right)-\left(z-1\right)\right]\left[\left(zx-x\right)-\left(z-1\right)\right]}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left[x\left(y-1\right)-\left(y-1\right)\right]\left[y\left(z-1\right)-\left(z-1\right)\right]\left[x\left(z-1\right)-\left(z-1\right)\right]}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-1\right)\left(y-1\right)\left(y-1\right)\left(z-1\right)\left(x-1\right)\left(z-1\right)}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-1\right)^2\left(y-1\right)^2\left(z-1\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-x-y-z\right)^2\left(y-x-y-z\right)^2\left(z-x-y-z\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(-y-z\right)^2\left(-x-z\right)^2\left(-x-y\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=1\)
Áp dụng BĐT AM-GM ta có:
\(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x\ge2\sqrt{9^x}=2\cdot3^x\)
\(\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge2\sqrt{25^x}=2\cdot5^x\)
\(\left(\frac{20}{3}\right)^x+\left(\frac{12}{5}\right)^x\ge2\sqrt{16^x}=2\cdot4^x\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left[\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\right]\ge2\left(3^x+4^x+5^x\right)\)
\(\Rightarrow\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge3^x+4^x+5^x\)
Trời thì ý bn là chứng minh bất đẳng thức côsi chứ j
Đây
Ta có: \(a,b\ge0\) nên \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)
Áp dụng hằng đẳng thức
Ta có: \(\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2-2\sqrt{a}\cdot\sqrt{b}\ge0\)
Suy ra \(a+b-2\sqrt{ab}\ge0\)
Suy ra \(a+b\ge2\sqrt{ab}\)và dấu ''='' xảy ra khi và chỉ khi a=b
Câu tiếp tương tự
Với lại hình như cái này lớp 7 đâu có học đâu mà hỏi nhỉ ????????
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\\\dfrac{a}{c}=\dfrac{b}{d}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left(\dfrac{a}{c}\right)^2=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\\\left(\dfrac{a}{c}\right)^2=\dfrac{ab}{cd}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
