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\(A=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\).Áp dụng BĐT Cauchy-Schwarz,ta có:
\(=\left(1-\frac{1}{x+1}\right)+\left(1-\frac{1}{y+1}\right)+\left(1-\frac{1}{z+1}\right)\)
\(=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)
Dấu "=" xảy ra khi x = y = z = 1/3
Vậy A min = 3/4 khi x=y=z=1/3
\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)
Ta có :
\(\frac{1+\sqrt{1+x^2}}{x}=\frac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\frac{2+\frac{4+1+x^2}{2}}{2x}=\frac{9+x^2}{4x}\)
tương tự : \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{9+y^2}{4y}\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{9+z^2}{4z}\)
\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le\frac{\left(9+x^2\right)yz+\left(9+y^2\right)xz+\left(9+z^2\right)xy}{4xyz}\)
\(=\frac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\frac{9\frac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=\frac{4\left(xyz\right)^2}{4xyz}=xyz\)
Dấu " = " xảy ra khi x = y = z = \(\sqrt{3}\)
\(\frac{x+1}{1+y^2}=\frac{\left(x+1\right)\left(y^2+1\right)-y^2\left(x+1\right)}{1+y^2}=x+1-\frac{y^2\left(x+1\right)}{1+y^2}\ge x+1-\frac{xy+y}{2}\)
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 \(x=y=z=1\)
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 \(P_{min}=3\Leftrightarrow x=y=z=1\)
Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)
\(A=\frac{1}{x^2+y^2}+\frac{3}{4xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{x^2+y^2+2xy}+1=\frac{4}{\left(x+y\right)^2}+1=5\)
Dấu "=" xảy ra khi x=y=1/2
Đúng ko biết !?
\(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}.\)
\(A=x^2y^2+\frac{1}{x^2y^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}=\left(x^2y^2+\frac{1}{256x^2y^2}\right)+\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+\frac{255}{256x^2y^2}\)
\(\ge2\sqrt{x^2y^2.\frac{1}{256x^2y^2}}+2\sqrt{\frac{x^2}{y^2}.\frac{y^2}{x^2}}+\frac{255}{256.\left(\frac{1}{4}\right)^2}=\frac{289}{16}.\)
\("="\text{ }for\text{ }x=y=\frac{1}{2}.\)
mk cx đâu có bít mới hok lớp 7 ak!!
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