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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 \(x=y=z=\frac{1}{3}\)
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 \(x=y=z=\sqrt{3}\)
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 \(x=y=z\)
Bài 1:
Áp dụng BĐT AM-GM cho các số thực dương ta có:
\(\frac{x^2}{y+z}+\frac{y+z}{4}\geq 2\sqrt{\frac{x^2}{4}}=x\)
\(\frac{y^2}{x+z}+\frac{x+z}{4}\geq 2\sqrt{\frac{y^2}{4}}=y\)
\(\frac{z^2}{x+y}+\frac{x+y}{4}\geq 2\sqrt{\frac{z^2}{4}}=z\)
Cộng theo vế:
\(\Rightarrow M+\frac{y+z}{4}+\frac{x+z}{4}+\frac{x+y}{4}\geq x+y+z\)
\(\Leftrightarrow M\geq \frac{x+y+z}{2}=\frac{2}{2}=1\)
Vậy GTNN của $M$ là $1$. Đẳng thức xảy ra tại $x=y=z=\frac{2}{3}$
Bài 2:
\(\text{VT}=(a+1)-\frac{b^2(a+1)}{b^2+1}+(b+1)-\frac{c^2(b+1)}{c^2+1}+(c+1)-\frac{a^2(c+1)}{a^2+1}\)
\(=(a+b+c+3)-\left(\frac{b^2(a+1)}{b^2+1}+\frac{c^2(b+1)}{c^2+1}+\frac{a^2(c+1)}{a^2+1}\right)\)
\(=6-M(*)\)
Xét \(M=\frac{b^2(a+1)}{b^2+1}+\frac{c^2(b+1)}{c^2+1}+\frac{a^2(c+1)}{a^2+1}\). Áp dụng BĐT AM-GM:
\(M\leq \frac{b^2(a+1)}{2b}+\frac{c^2(b+1)}{2c}+\frac{a^2(c+1)}{2a}=\frac{ab+bc+ac+a+b+c}{2}=\frac{ab+bc+ac+3}{2}\)
\(\leq \frac{\frac{(a+b+c)^2}{3}+3}{2}=3(**)\)
Từ \((*); (**)\Rightarrow \text{VT}=6-M\geq 6-3=3\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
Áp dụng BĐT AM-GM ta có:
\(\dfrac{x+1}{1+y^2}=x+1-\dfrac{y^2\left(x+1\right)}{y^2+1}\ge x+1-\dfrac{y\left(x+1\right)}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y+1}{1+z^2}\ge y+1-\dfrac{z\left(y+1\right)}{2};\dfrac{z+1}{1+x^2}\ge z+1-\dfrac{x\left(z+1\right)}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(Q\ge\left(x+y+z+3\right)-\dfrac{x\left(z+1\right)+y\left(x+1\right)+z\left(y+1\right)}{2}\)
\(=6-\dfrac{xy+yz+xz+x+y+z}{2}\)
\(\ge6-\dfrac{\dfrac{\left(x+y+z\right)^2}{3}+3}{2}=6-3=3\)
Đẳng thức xảy ra khi \(x=y=z=1\)
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) = \(\dfrac{3}{4}\)
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 \(x=\dfrac{1}{6};y=\dfrac{1}{3};z=\dfrac{1}{2}\)