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\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\)
Ta có:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge\frac{3a}{4}\)
\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\frac{6a-b-c-2}{8}\)
Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(1+c\right)\left(1+a\right)}\ge\frac{6b-c-a-2}{8}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6c-a-b-2}{8}\end{cases}}\)
Cộng vế theo vế ta được
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6a-b-c-2}{8}+\frac{6b-c-a-2}{8}+\frac{6c-a-b-2}{8}\)
\(=\frac{a+b+c}{2}-\frac{3}{4}\ge\frac{3}{2}.\sqrt[3]{abc}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)
\(ab+bc+ca\le a^2+b^2+c^2\le\frac{\left(a+b+c\right)^2}{3}\) ( bđt phụ + Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
CM bđt phụ : \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Leftrightarrow\)\(2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)
\(\Leftrightarrow\)\(2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(\Leftrightarrow\)\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)
\(\Leftrightarrow\)\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) ( luôn đúng )
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)
Chúc bạn học tốt ~
Áp dụng bất đẳn thức Cauchy-Schwarz ta có:
\(\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)\left(x+y+z\right)=\)\(\left[\frac{a^2}{\left(\sqrt{x}\right)^2}+\frac{b^2}{\left(\sqrt{y}\right)^2}+\frac{c^2}{\left(\sqrt{z}\right)^2}\right]\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)
\(\ge\left(\frac{a}{\sqrt{x}}.\sqrt{x}+\frac{b}{\sqrt{y}}.\sqrt{y}+\frac{c}{\sqrt{z}}.\sqrt{z}\right)=\left(a+b+c\right)\)\(\left(đpcm\right)\)
Bài 1:
Ta có: \(P=\frac{1}{1+x^2}+\frac{4}{4+y^2}=\frac{1}{1+x^2}+\frac{1}{1+\frac{y^2}{4}}\)
Đặt \(\left(x;\frac{y}{2}\right)=\left(a;b\right)\left(a,b>0\right)\)
\(\Rightarrow\hept{\begin{cases}P=\frac{1}{1+a^2}+\frac{1}{1+b^2}+2ab\\ab\ge1\end{cases}}\)
Ta có: \(P=\frac{1}{1+a^2}+\frac{1}{1+b^2}+2ab\)
\(\ge\frac{1}{ab+a^2}+\frac{1}{ab+b^2}+2ab=\frac{1}{ab}+2ab\)
\(=\left(\frac{1}{ab}+ab\right)+ab\ge2+1=3\)
Dấu "=" xảy ra khi: \(ab=\frac{1}{ab}\Rightarrow ab=1\Rightarrow xy=2\)
Bài 3:
Đặt \(\left(a-1;b-1;c-1\right)=\left(x;y;z\right)\left(x,y,z>1\right)\)
Khi đó:
\(BĐTCCM\Leftrightarrow\frac{\left(x+1\right)^2}{y}+\frac{\left(y+1\right)^2}{z}+\frac{\left(z+1\right)^2}{x}\ge12\)
Thật vậy vì ta có:
\(VT=\frac{\left(x+1\right)^2}{y}+\frac{\left(y+1\right)^2}{z}+\frac{\left(z+1\right)^2}{x}\)
\(=\frac{x^2+2x+1}{y}+\frac{y^2+2y+1}{z}+\frac{z^2+2z+1}{x}\)
\(=\left(\frac{2x}{y}+\frac{2y}{z}+\frac{2z}{x}\right)+\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Áp dụng BĐT Cauchy ta có:
\(VT\ge3\sqrt[3]{\frac{2x}{y}\cdot\frac{2y}{z}\cdot\frac{2z}{x}}+6\sqrt[6]{\frac{x^2}{y}\cdot\frac{y^2}{z}\cdot\frac{z^2}{x}\cdot\frac{1}{x}\cdot\frac{1}{y}\cdot\frac{1}{z}}=6+6=12\)
Dấu "=" xảy ra khi: \(x=y=z\Leftrightarrow a=b=c\)
a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)
b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)
\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)
Câu 2/
\(\frac{a^2+bc}{a^2\left(b+c\right)}+\frac{b^2+ca}{b^2\left(c+a\right)}+\frac{c^2+ab}{c^2\left(a+b\right)}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{a^2+bc}{a^2\left(b+c\right)}-\frac{1}{a}+\frac{b^2+ca}{b^2\left(c+a\right)}-\frac{1}{b}+\frac{c^2+ab}{c^2\left(a+b\right)}-\frac{1}{c}\ge0\)
\(\Leftrightarrow\frac{\left(b-a\right)\left(c-a\right)}{a^2\left(b+c\right)}+\frac{\left(a-b\right)\left(c-b\right)}{b^2\left(c+a\right)}+\frac{\left(a-c\right)\left(b-c\right)}{c^2\left(a+b\right)}\ge0\)
\(\Leftrightarrow a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-a^2b^4c^2-a^2b^2c^4\ge0\)
\(\Leftrightarrow a^4b^4+b^4c^4+c^4a^4\ge a^4b^2c^2+a^2b^4c^2+a^2b^2c^4\left(1\right)\)
Ma ta có: \(\hept{\begin{cases}a^4b^4+b^4c^4\ge2a^2b^4c^2\left(2\right)\\b^4c^4+c^4a^4\ge2a^2b^2c^4\left(3\right)\\c^4a^4+a^4b^4\ge2a^4b^2c^2\left(4\right)\end{cases}}\)
Cộng (2), (3), (4) vế theo vế rồi rút gọn cho 2 ta được điều phải chứng minh là đúng.
PS: Nếu nghĩ được cách khác đơn giản hơn sẽ chép lên cho b sau. Tạm cách này đã.
tks bn nhé, bn giúp mk câu 1 được ko