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ĐK: \(x\ge-1;y\ge3;z\ge1\)
\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}=\dfrac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow x+1-2\sqrt{x+1}+1+y-3-2\sqrt{y-3}+1+z-1-2\sqrt{z-1}+1=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)
Ta thấy: \(\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2\ge0\)
Đẳng thức xảy ra khi:
\(\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-3}=1\\\sqrt{z-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)
Cách khác:
ĐK: \(x\ge-1;y\ge3;z\ge1\)
Áp dụng BĐT \(ab\le\dfrac{a^2+b^2}{2}\).
\(\sqrt{x+1}\le\dfrac{x+1+1}{2}=\dfrac{x+2}{2}\)
\(\sqrt{y-3}\le\dfrac{y-3+1}{2}=\dfrac{y-2}{2}\)
\(\sqrt{z-1}\le\dfrac{z-1+1}{2}=\dfrac{z}{2}\)
Cộng vế theo vế các BĐT trên ta được:
\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}\le\dfrac{1}{2}\left(x+y+z\right)\)
Đẳng thức xảy ra khi:
\(\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-3}=1\\\sqrt{z-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)
ĐKXĐ: \(x\ge-1;y\ge3;z\ge1\)
\(\Leftrightarrow x+y+z-2\sqrt{x+1}-2\sqrt{y-3}-2\sqrt{z-1}=0\)
\(\Leftrightarrow\left(x+1-2\sqrt{x+1}+1\right)+\left(y-3-2\sqrt{y-3}+1\right)+\left(z-1-2\sqrt{z-1}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}-1=0\\\sqrt{y-3}-1=0\\\sqrt{z-1}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)
Lời giải:
Từ \(x+y+z=xyz\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
Đặt \((\frac{1}{a}, \frac{1}{b}, \frac{1}{c})=(x,y,z)\), trong đó $a,b,c>0$ thì ta có:
\(ab+bc+ac=1\) và cần phải CMR:
\(P=\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}+\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}+\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}\)
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Ta có:
\(\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}=\sqrt{(b^2+1)(c^2+1)}-b\sqrt{c^2+1}-c\sqrt{b^2+1}\)
\(=\sqrt{(b^2+ab+bc+ac)(c^2+ac+bc+ab)}-b\sqrt{c^2+ac+bc+ab}-c\sqrt{b^2+ab+bc+ac}\)
\(=\sqrt{(b+a)(b+c)(c+a)(c+b)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}\)
\(=(b+c)\sqrt{(a+b)(a+c)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}(1)\)
Tương tự:
\(\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}=(a+c)\sqrt{(b+a)(b+c)}-a\sqrt{(c+a)(c+b)}-c\sqrt{(a+b)(a+c)}(2)\)
\(\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}=(a+b)\sqrt{(c+a)(c+b)}-b\sqrt{(a+b)(a+c)}-a\sqrt{(b+c)(b+a)}(3)\)
Từ \((1);(2);(3)\Rightarrow P=(b+c-c-b)\sqrt{(a+b)(a+c)}+(a+c-c-a)\sqrt{(b+a)(b+c)}+(a+b-b-a)\sqrt{(c+a)(c+b)}\)
\(=0\)
Ta có đpcm.
\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
= có x +y+z=a=>x2+y2+z2+2(xy+yz+xz)=a2
Thay vào a2=b+3992=>xy+zy+xz=1996
thay vào P ta có
P=x\(\sqrt{\dfrac{\left(xy+yz+zx+z^2\right)\left(zx+xy+yz+x^2\right)}{xy+yz+zx+x^2}}\)
+y\(\sqrt{\dfrac{\left(zx+zy+xy+z^2\right)\left(zx+zy+xy+x^2\right)}{xy+yz+xz+y^2}}\)
+\(\sqrt{\dfrac{\left(zx+xy+zy+x^2\right)\left(xz+xy+zy+y^2\right)}{xz+xy+zy+z^2}}\)
=x\(\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}\)
+y\(\sqrt{\dfrac{\left(x+z\right)\left(z+y\right)\left(x+y\right)\left(z+x\right)}{\left(y+z\right)\left(x+y\right)}}\)
+z\(\sqrt{\dfrac{\left(y+z\right)\left(x+y\right)\left(z+x\right)\left(x+y\right)}{\left(z+x\right)\left(z+y\right)}}\)
=x\(\sqrt{\left(y+z\right)^2}\)+y\(\sqrt{\left(x+z\right)^2}\)+z\(\sqrt{\left(x+y\right)^2}\)=x(z+y)+y(x+z)+z(x+y)
=2(xy+zx+zy)=3992
*có gì ko hiểu thì hỏi
