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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\geq \frac{(x+y+z)^2}{x+y+y+z+z+x}\)
\(\Leftrightarrow A\geq \frac{x+y+z}{2}\)
Áp dụng BĐT AM-GM:
\(\left\{\begin{matrix} x+y\geq 2\sqrt{xy}\\ y+z\geq 2\sqrt{yz}\\ z+x\geq 2\sqrt{zx}\end{matrix}\right.\)
\(\Rightarrow 2(x+y+z)\geq 2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})=2\)
\(\Rightarrow x+y+z\geq 1\)
Do đó: \(A\geq \frac{x+y+z}{2}\geq \frac{1}{2}\)
Vậy \(A_{\min}=\frac{1}{2}\)
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)
Lời giải:
Do \(x+y+z=1\) nên biến đổi như sau:
\(P=\frac{x}{(x+y)+(x+z)}+\frac{y}{(y+z)+(y+x)}+\frac{z}{(z+x)+(z+y)}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{(x+y)+(x+z)}\leq \frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\Rightarrow \frac{x}{(x+y)+(x+z)}\leq \frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Thực hiện tương tự với các phân thức còn lại:
\(\Rightarrow P\leq \frac{1}{4}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{x+z}\right)=\frac{3}{4}\)
Vậy \(P_{\max}=\frac{3}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)
\(P=\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
Thay \(x+y+z=1\) vào biểu thức
\(\Rightarrow P=\dfrac{x}{2x+y+z}+\dfrac{y}{x+2y+z}+\dfrac{z}{x+y+2z}\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2x+y+z}=\dfrac{x}{x+y+x+z}\le\dfrac{x}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\\\dfrac{y}{x+2y+z}=\dfrac{y}{x+y+y+z}\le\dfrac{y}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\\\dfrac{z}{x+y+2z}=\dfrac{z}{x+z+y+z}\le\dfrac{z}{4}\left(\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{x}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)+\dfrac{y}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)+\dfrac{z}{4}\left(\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)
\(\Rightarrow VT\le\dfrac{x}{4\left(x+y\right)}+\dfrac{x}{4\left(x+z\right)}+\dfrac{y}{4\left(x+y\right)}+\dfrac{y}{4\left(y+z\right)}+\dfrac{z}{4\left(x+z\right)}+\dfrac{z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\dfrac{x}{4\left(x+y\right)}+\dfrac{y}{4\left(x+y\right)}+\dfrac{x}{4\left(x+z\right)}+\dfrac{z}{4\left(x+z\right)}+\dfrac{y}{4\left(y+z\right)}+\dfrac{z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\dfrac{x+y}{4\left(x+y\right)}+\dfrac{x+z}{4\left(x+z\right)}+\dfrac{y+z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Rightarrow P\le\dfrac{3}{4}\)
Vậy \(P_{max}=\dfrac{3}{4}\)
Dấu '' = '' xảy ra khi \(x=y=z\)
\(l=\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}=\dfrac{1^2}{x}+\dfrac{2^2}{y}+\dfrac{3^2}{z}\ge\dfrac{\left(1+2+3\right)^2}{x+y+z}=\dfrac{36}{1}=36\)
ta có:\(P=\sum\dfrac{y^2z^2}{x\left(y^2+z^2\right)}=\sum\dfrac{\dfrac{1}{x}}{\dfrac{1}{y^2}+\dfrac{1}{z^2}}\)
đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\)thì giả thiết trở thành : \(a^2+b^2+c^2=1\).tìm Min \(P=\dfrac{a}{b^2+c^2}+\dfrac{b}{a^2+c^2}+\dfrac{c}{a^2+b^2}\)
ta có:\(\dfrac{a}{b^2+c^2}=\dfrac{a}{1-a^2}=\dfrac{a^2}{a\left(1-a^2\right)}\)
Áp dụng bất đẳng thức cauchy:
\(\left[a\left(1-a^2\right)\right]^2=\dfrac{1}{2}.2a^2\left(1-a^2\right)\left(1-a^2\right)\le\dfrac{1}{54}\left(2a^2+1-a^2+1-a^2\right)^3=\dfrac{4}{27}\)
\(\Rightarrow a\left(1-a^2\right)\le\dfrac{2}{3\sqrt{3}}\)\(\Rightarrow\dfrac{a^2}{a\left(1-a^2\right)}\ge\dfrac{3\sqrt{3}}{2}a^2\)
tương tự với các phân thức còn lại ta có:
\(P\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
hay \(x=y=z=\sqrt{3}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\\\dfrac{1}{z}=c\end{matrix}\right.\) Thì bài toán trở thành
Cho \(a^2+b^2+c^2=1\) tính GTNN của \(P=\dfrac{a}{b^2+c^2}+\dfrac{b}{c^2+a^2}+\dfrac{c}{a^2+b^2}\)
Ta có:
\(a^2+b^2+c^2=1\)
\(\Rightarrow a^2+b^2=1-c^2\)
\(\Rightarrow\dfrac{c}{a^2+b^2}=\dfrac{c^2}{c\left(1-c^2\right)}\)
Mà ta có: \(2c^2\left(1-c^2\right)\left(1-c^2\right)\le\dfrac{\left(2c^2+1-c^2+1-c^2\right)^3}{27}=\dfrac{8}{27}\)
\(\Rightarrow c\left(1-c^2\right)\le\dfrac{2}{3\sqrt{3}}\)
\(\Rightarrow\dfrac{c^2}{c\left(1-c^2\right)}\ge\dfrac{3\sqrt{3}c^2}{2}\)
\(\Rightarrow\dfrac{c}{a^2+b^2}\ge\dfrac{3\sqrt{3}c^2}{2}\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{b}{c^2+a^2}\ge\dfrac{3\sqrt{3}b^2}{2}\left(2\right)\\\dfrac{a}{b^2+c^2}\ge\dfrac{3\sqrt{3}a^2}{2}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) \(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)
Áp dụng bđt Cauchy-Schwarz:
\(\frac{1}{x}+\frac{9}{y}+\frac{16}{z}\ge\frac{\left(1+3+4\right)^2}{x+y+z}=64\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\frac{1}{x}=\frac{3}{y}=\frac{4}{z}\\x+y+z=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{8}\\y=\frac{3}{8}\\z=\frac{1}{2}\end{matrix}\right.\)
bài 3:
a, đặt x12=y9=z5=kx12=y9=z5=k
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: x5=y7=z3=x225=y249=z29x5=y7=z3=x225=y249=z29
A/D tính chất dãy tỉ số bằng nhau ta có:
x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
Áp dụng bđt Bunhiacopxki ta có :
\(A=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\right)\ge\left(\sqrt{x}.\dfrac{1}{\sqrt{x}}+\sqrt{y}.\dfrac{2}{\sqrt{y}}+\sqrt{z}.\dfrac{3}{\sqrt{z}}\right)^2\)
\(\left(1+2+3\right)^2=36\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel
\(A\ge\dfrac{\left(1+2+3\right)^2}{x+y+z}=36\)
Đẳng thức xảy ra khi \(x=\dfrac{1}{6};y=\dfrac{1}{3};z=\dfrac{1}{2}\)