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a) Chứng minh được BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(*)
Dấu "=" xảy ra <=> a=b
Áp dụng BĐT (*) vào bài toán ta có:
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+y}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{1}{x+y+2z}=\frac{1}{x+y+z+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Tiếp tục áp dụng BĐT (*) ta có:
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right);\frac{1}{y+z}\le\frac{1}{4}\left(\frac{1}{y}+\frac{1}{z}\right);\frac{1}{z+x}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\cdot\frac{1}{4}\cdot2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)
b) áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có:
\(\hept{\begin{cases}\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{b+c-a+a+c-b}=\frac{4}{2c}=\frac{2}{c}\\\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{a+b-c+a+c-b}=\frac{4}{2a}=\frac{2}{a}\end{cases}}\)
Cộng theo vế 3 BĐT ta có:
\(2VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VP\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra <=> a=b=c
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
\(a+b+c+ab+ac+bc=6abc\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\) \(\Rightarrow x+y+z+xy+xz+yz=6\)
Cần chứng minh \(P=x^2+y^2+z^2\ge3\)
Ta có các BĐT quen thuộc:
\(x^2+1\ge2x\) ; \(y^2+1\ge2y\); \(z^2+1\ge2z\)
\(2x^2+2y^2+2z^2\ge2xy+2xz+2yz\)
Cộng vế với vế:
\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+xz+yz\right)=12\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge9\)
\(\Rightarrow x^2+y^2+z^2\ge3\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
Vì a+b+c=1 nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{a}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)=2+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\)
Do đó
\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\left(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{ab}\right)+\left(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{bc}\right)+\left(\frac{ca}{a^2+c^2}+\frac{c^2+a^2}{ca}\right)+\frac{3}{4}\)
\(\ge2\sqrt{\frac{ab}{a^2+b^2}\cdot\frac{a^2+b^2}{ab}}+2\sqrt{\frac{bc}{c^2+b^2}\cdot\frac{c^2+b^2}{bc}}+2\sqrt{\frac{ca}{a^2+c^2}+\frac{c^2+a^2}{ca}}+\frac{3}{4}\)
\(=2\cdot\frac{1}{2}+2\cdot\frac{1}{2}+\frac{2}{3}=\frac{15}{4}\)
Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)
\(a+b+c+ab+bc+ca=6abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z+xy+yz+zx=6\)
Ta cần chứng minh: \(x^2+y^2+z^2\ge3\)
Thật vậy:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)
Cộng vế với vế:
\(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge12\)
\(\Rightarrow x^2+y^2+z^2\ge3\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;1\right)\) hay \(\left(a;b;c\right)=\left(1;1;1\right)\)
Áp dụng BĐT cho 2 số dương:
\(\frac{1}{\left(a+b\right)}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Xét: c + 1 = c + a + b + c
\(\frac{ab}{\left(c+1\right)}\le\frac{ab}{4}.\left[\frac{1}{\left(a+c\right)}+\frac{1}{\left(b+c\right)}\right]\)
Tương tự:
\(\frac{bc}{\left(a+1\right)}\le\frac{bc}{4}.\left[\frac{1}{\left(a+c\right)}+\frac{1}{\left(b+a\right)}\right]\)
\(\frac{ca}{\left(b+1\right)}\le\frac{ac}{4}.\left[\frac{1}{\left(a+b\right)}+\frac{1}{\left(c+b\right)}\right]\)
Cộng lại:
\(\frac{ac}{\left(c+1\right)}+\frac{bc}{\left(a+1\right)}+\frac{ca}{\left(b+1\right)}\le\frac{1}{4}\left\{\frac{ab}{\left(a+c\right)}+\frac{ab}{\left(b+c\right)}+\frac{bc}{\left(a+c\right)}+\frac{bc}{\left(a+c\right)}+\frac{ac}{\left(a+b\right)}\right\}\)
Cộng lại + rút gọn mẫu số
\(\frac{ab}{\left(c+1\right)}+\frac{bc}{\left(a+1\right)}+\frac{ca}{b+1}\le\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)
Dấu '=' xảy ra khi a = b = c
P/s: Sai đâu bạn sửa nhé!