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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\\ \Leftrightarrow\dfrac{x+y}{xy}+\left(\dfrac{1}{z}-\dfrac{1}{x+y+z}\right)=0\\ \Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y}{z\left(x+y+z\right)}=0\\ \Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}\right)=0\\ \)
Nếu x+y=0 => x=-y
Nếu
\(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}=0\\ \Rightarrow xz+yz+z^2+xy=0\\ \Rightarrow\left(x+z\right)\left(y+z\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-z\\y=-z\end{matrix}\right.\)
Tự thế vào :v
Từ \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)
\(\Rightarrow\dfrac{x+y}{xy}+\dfrac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
Ta có: x8 - y8 = (x + y)(x - y)(x2 + y2)(x4 + y4)
y9 + z9 = (y + z)(y8 - y7z + y6z2 - ... + z8)
z10 - x10 = (z + x)(z4 - z3x + z2x2 - zx3 + z4)(z5 - x5)
Vậy M = \(\dfrac{3}{4}\) + (x + y)(y + z)(z + x) = \(\dfrac{3}{4}\)
Ta có \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\Rightarrow\left(xy+xz+yz\right)\left(x+y+z\right)=xyz\Rightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\Rightarrow\left[{}\begin{matrix}x+y=0\\x+z=0\\y+z=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-y\\z=-x\\y=-z\end{matrix}\right.\)TH1: Nếu x=-y⇒x8-y8=x8-(-x)8=0 (Vì x8 và (-x)8 đều là số nguyên dương)⇒M=\(\text{}\dfrac{3}{4}+\left(x^8-y^8\right)\left(y^9-z^9\right)\left(z^{10}-x^{10}\right)=\dfrac{3}{4}\)
Tương tự với y=-z và z=-x
Vậy M=\(\dfrac{3}{4}\)
Lời giải:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
$\Rightarrow (\frac{1}{x}+\frac{1}{y})+(\frac{1}{z}-\frac{1}{x+y+z})=0$
$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$
$\Leftrightarrow (x+y)(\frac{1}{xy}+\frac{1}{z(x+y+z)})=0$
$\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y)(y+z)(x+z)=0$
$\Leftrightarrow x=-y$ hoặc $y=-z$ hoặc $z=-x$
Nếu $x=-y$ thì:
$P=\frac{3}{4}+[(-y)^8-y^8](y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}+0.(y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}$
Nếu $y=-z$ thì:
$P=\frac{3}{4}+(x^8-y^8)[(-z)^9+z^9](z^{10}-x^{10})=\frac{3}{4}+(x^8-y^8).0.(z^{10}-x^{10})=\frac{3}{4}$
Nếu $z=-x$ thì:
$P=\frac{3}{4}+(x^8-y^8)(y^9+z^9)[(-x)^{10}-x^{10}]=\frac{3}{4}+(x^8-y^8)(y^9+z^9).0=\frac{3}{4}$
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow x=-y\text{ hoặc }y=-z\text{ hoặc }z=-x\)
\(+\text{Nếu }x=-y\text{ thì }x^8=\left(-y\right)^8=y^8\Rightarrow x^8-y^8=0\Rightarrow M=\frac{3}{4}\)
\(+\text{Nếu }y=-z\text{ thì }y^9=\left(-z\right)^9=-z^9\Rightarrow y^9+z^9=0\Rightarrow M=\frac{3}{4}\)
\(+\text{Nếu }z=-x\text{ thì }z^{10}=\left(-x\right)^{10}=x^{10}\Rightarrow z^{10}-x^{10}=0\Rightarrow M=\frac{3}{4}\)
\(\text{Vậy M}=\frac{3}{4}.\)
\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=8\)
=>\(8xyz=xyz+\sum x+\sum xy+1\)
=>\(\sum x^2+14xyz=\left(\sum x\right)^2+2\sum x+2\)
mặt khác
\(8=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\ge\dfrac{8}{\sqrt[3]{xyz}}\rightarrow xyz\ge1\)
đặt \(\sum x=a\left(a\ge3\right)\)
khi đó \(P=\dfrac{a^2+2a+2}{4a^2+15xyz}\le\dfrac{a^2+2a+2}{4a^2+15}\)
\(\dfrac{a^2+2a+2}{4a^2+15}=\dfrac{1}{3}-\dfrac{\left(a-3\right)^2}{12a^2+45}\le\dfrac{1}{3}\)
vậy max bằng 1/3 khi x=y=z=1
Bài 1:
\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )
Khi đó:
\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)
\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)
(Áp dụng BĐT Cauchy_Schwarz)
Theo BĐT Cauchy dễ thấy:
\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$
Bài 2:
Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)
Ta có:
\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)
\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)
Áp dụng BĐT Cauchy:
\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)
\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)
\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)
Nhân theo vế:
\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)
\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)
\(\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$
ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)
\(\Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(\dfrac{xz+yz+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\x+z=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^8=\left(-y\right)^8\\y^9=\left(-z\right)^9\\z^{10}=\left(-x\right)^{10}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^8-y^8=0\\y^9+z^9=0\\x^{10}-z^{10}=0\end{matrix}\right.\)\(\Rightarrow\left(x^8-y^8\right)\left(y^9+z^9\right)\left(z^{10}-x^{10}\right)=0\)
\(\Rightarrow M=\dfrac{3}{4}\)