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\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)
Ta có: \(\frac{x+1}{y^2+1}=\left(x+1\right).\frac{1}{y^2+1}=\left(x+1\right)\left(1-\frac{y^2}{y^2+1}\right)\)
\(\ge\left(x+1\right)\left(1-\frac{y^2}{2y}\right)=x+1-\frac{y\left(x+1\right)}{2}\)
Thiết lập hai BĐT còn lại tương tự và cộng theo vế:
\(P\ge\left(x+y+z+3\right)-\frac{x\left(z+1\right)+y\left(x+1\right)+z\left(y+1\right)}{2}\)
\(=6-\frac{\left(xy+yz+zx\right)+\left(x+y+z\right)}{2}\) (*)
Lại có BĐT \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)
Thật vậy,ta có: BĐT \(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng)
Thay vào (*),ta có: \(P\ge6-\frac{\left(xy+yz+zx\right)+\left(x+y+z\right)}{2}\)
\(\ge6-\frac{\frac{\left(x+y+z\right)^2}{3}+3}{2}=6-\frac{3+3}{2}=3\)
Dấu "=" xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)
Vậy \(P_{min}=3\Leftrightarrow x=y=z=1\)
\(\frac{x+1}{1+y^2}=\frac{\left(x+1\right)\left(y^2+1\right)-y^2\left(x+1\right)}{1+y^2}=x+1-\frac{y^2\left(x+1\right)}{1+y^2}\ge x+1-\frac{xy+y}{2}\)
Tương tự ta có:
\(\frac{y+1}{z^2+1}\ge y+1-\frac{yz+z}{2}\)
\(\frac{z+1}{1+x^2}\ge z+1-\frac{zx+x}{2}\)
Cộng vế theo vế ta có:
\(Q\ge3+\left(x+y+z\right)-\frac{x+y+z+xy+yz+zx}{2}\)
\(=3+\frac{x+y+z-xy-yz-zx}{2}\)
Có BĐT phụ sau:
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) ( tự cm )
\(\Rightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=3\)
Khi đó \(P\ge3\)
Dấu "=" xảy ra tại \(x=y=z=1\)
Cay, đánh xong rồi tự nhiên bấm hủy :v
Ta có:\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)
Khi đó:
\(A=\frac{a^2\left(1+2b\right)}{b}+\frac{b^2\left(1+2c\right)}{c}+\frac{c^2\left(1+2a\right)}{a}\)
\(=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+2\left(a^2+b^2+c^2\right)\)
\(\ge\frac{\left(a+b+c\right)^2}{a+b+c}+2\cdot\frac{\left(a+b+c\right)^2}{3}\)
\(=a+b+c+\frac{2\left(a+b+c\right)^2}{3}\)
\(\ge\sqrt{3\left(ab+bc+ca\right)}+\frac{6\left(ab+bc+ca\right)}{3}\)
\(=2+\sqrt{3}\)
Đẳng thức xảy ra tại \(x=y=z=\sqrt{3}\)
zZz Cool Kid_new zZz. Sai đề rồi bạn êii !
Nếu bạn đặt như vậy thì
\(A=\frac{y-2}{x^2}+\frac{z-2}{y^2}+\frac{x-2}{z^2}\)
\(=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}\)
\(=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-2.\left(a^2+b^2+c^2\right)\)
Ta có: \(x+y+z=xyz\Rightarrow x=\frac{x+y+z}{yz}\Rightarrow x^2=\frac{x^2+xy+xz}{yz}\Rightarrow x^2+1=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)\(\Rightarrow\sqrt{x^2+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}\le\frac{2+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự: \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3.\frac{xy+yz+zx}{xyz}\)\(\le3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}=\frac{\left(xyz\right)^2}{xyz}=xyz\)
Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)
\(P=\dfrac{2a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}=\dfrac{2a}{\sqrt{ab+bc+ca+a^2}}+\dfrac{b}{\sqrt{ab+bc+ca+b^2}}+\dfrac{c}{\sqrt{ab+bc+ca+c^2}}\)
\(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(P=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{c+a}.\dfrac{c}{2\left(c+b\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{c+a}+\dfrac{c}{2\left(c+b\right)}\right)=\dfrac{9}{4}\)
\(P_{max}=\dfrac{9}{4}\) khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\) hay \(\left(x;y;z\right)=\left(\dfrac{\sqrt{15}}{7};\sqrt{15};\sqrt{15}\right)\)
Ta có \(x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2=4\Rightarrow+xy+yz+zx=-7\)
vì \(x+y+z=2\Rightarrow z-1=1-x-y\Rightarrow\frac{1}{xy+z-1}=\frac{1}{xy+1-x-y}=\frac{1}{\left(x-1\right)\left(y-1\right)}. \)
Suy ra \(S=\frac{1}{\left(x-1\right)\left(y-1\right)}+\frac{1}{\left(y-1\right)\left(z-1\right)}+\frac{1}{\left(z-1\right)\left(x-1\right)}. \)
\(\frac{z-1+x-1+y-1}{\left(x-1\right)\left(y-1\right)\left(z-1\right)}=\frac{x+y+z-3}{xyz-xy-yz-zx+x+y+z-1}=-\frac{1}{7}\)