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Đặt \(b+c-a=2x;c+a-b=2y;a+b-c=2z\) \(\Rightarrow a=y+z;b=x+z;c=x+y\)
\(\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}=\dfrac{4\left(y+z\right)}{2x}+\dfrac{4\left(x+z\right)}{2y}+\dfrac{4\left(x+y\right)}{2z}\)\(=\dfrac{2\left(y+z\right)}{x}+\dfrac{2\left(x+z\right)}{y}+\dfrac{2\left(x+y\right)}{z}=2\left(\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{x}{y}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{y}{z}\right)\ge2.\left(2+2+2\right)=12\)
Đặt \(b+c-a=2x;c+a-b=2y;a+b-c=2z\)\(\Rightarrow a=y+z;b=z+x;c=x+y\)
\(P=\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}=\dfrac{4\left(y+z\right)}{2x}+\dfrac{4\left(z+x\right)}{2y}+\dfrac{4\left(x+y\right)}{2z}\)\(\Leftrightarrow\dfrac{2\left(y+z\right)}{x}+\dfrac{2\left(z+x\right)}{y}+\dfrac{2\left(x+y\right)}{z}=2\left(\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{z}\right)\ge2.\left(2+2+2\right)=12\)
dạng này chắc chắc là phải dùng AM-GM ngược dấu rồi :)
Ta có:
\(\dfrac{1+b}{1+4a^2}=1+b-\dfrac{4a^2\left(b+1\right)}{4a^2+1}\ge1+b-\dfrac{4a^2\left(b+1\right)}{4a}=1+b-a\left(b+1\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\dfrac{1+c}{1+4b^2}\ge1+c-b\left(c+1\right);\dfrac{1+a}{1+4c^2}\ge1+a-c\left(a+1\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT=\dfrac{1+b}{1+4a^2}+\dfrac{1+c}{1+4b^2}+\dfrac{1+a}{1+c^2}\)
\(\ge3+\left(a+b+c\right)-\left(ab+bc+ca\right)-\left(a+b+c\right)\)
\(=3-\dfrac{1}{3}\left(a+b+c\right)^2=3-\dfrac{1}{3}\cdot\dfrac{9}{4}=\dfrac{9}{4}=VP\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{2}\)
\(VT=\left(\dfrac{a}{1+4c^2}+\dfrac{b}{1+4a^2}+\dfrac{c}{1+4b^2}\right)+\left(\dfrac{1}{1+4c^2}+\dfrac{1}{1+4a^2}+\dfrac{1}{1+4b^2}\right)\)
\(VT=\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)+3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\)
Xét \(\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+4c^2\ge2\sqrt{4c^2}=4c\\1+4a^2\ge2\sqrt{4a^2}=4a\\1+4b^2\ge2\sqrt{4b^2}=4b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4c^2a}{1+4c^2}\le\dfrac{4c^2a}{4c}=ca\\\dfrac{4a^2b}{1+4a^2}\le\dfrac{4a^2b}{4a}=ab\\\dfrac{4b^2c}{1+4b^2}\le\dfrac{4b^2c}{4b}=bc\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)\ge\dfrac{3}{2}-\left(ab+bc+ca\right)\) (1)
Xét \(3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+4c^2\ge2\sqrt{4c^2}=4c\\1+4a^2\ge2\sqrt{4a^2}=4a\\1+4b^2\ge2\sqrt{4b^2}=4b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4c^2}{1+4c^2}\le\dfrac{4c^2}{4c}=c\\\dfrac{4a^2}{1+4a^2}\le\dfrac{4a^2}{4a}=a\\\dfrac{4b^2}{1+4b^2}\le\dfrac{4b^2}{4b}=b\end{matrix}\right.\)
\(\Rightarrow3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\ge\dfrac{3}{2}\) (2)
Từ (1) và (2)
\(\Rightarrow VT\ge\dfrac{3}{2}-\left(ab+bc+ca\right)+\dfrac{3}{2}\)
\(\Rightarrow VT\ge3-\left(ab+bc+ca\right)\) (3)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{3}{4}\ge ab+bc+ca\)
\(\Rightarrow3-\dfrac{3}{4}\le3-\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{9}{4}\le3-\left(ab+bc+ca\right)\) (4)
Từ (3) và (4)
\(\Rightarrow VT\ge\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{1+b}{1+4a^2}+\dfrac{1+c}{1+4b^2}+\dfrac{1+a}{1+4c^2}\ge\dfrac{9}{4}\) (đpcm)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{2}\)
1,
\(x^2+y^2+y^2=14\)
\(\Rightarrow\left(x+y+z\right)^2-2xy-2yz-2zx=14\)
\(\Rightarrow-2\left(xy+yz+zx\right)=14\)
\(\Rightarrow xy+yz+zx=-7\)
\(\Rightarrow\left(xy+yz+zx\right)^2=49\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2x^2yz+2xy^2z+2xyz^2=49\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)=49\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2=49\)
Ta có: \(x^4+y^4+z^4\)
\(=\left(x^2+y^2+z^2\right)^2-2x^2y^2-2y^2z^2-2z^2x^2\)
\(=14^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)\)
\(=14^2-2.49\)
\(=196-98\)
\(=98\)
\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)