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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 $x=y=z=1$
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 \(x=\dfrac{1}{6};y=\dfrac{1}{3};z=\dfrac{1}{2}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+x+y+z)\geq (1+1+1+1)^2\)
\(\Rightarrow \frac{2}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{16}{2x+y+z}\)
Hoàn toàn tương tự:
\(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\geq \frac{16}{x+2y+z}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\geq \frac{16}{x+y+2z}\)
Cộng theo vế các BĐT vừa thu được:
\(\Rightarrow 4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(\Rightarrow 16\geq 16\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(\Rightarrow \frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\leq 1\)
Ta có đpcm.
Ta có :
\(\dfrac{1}{2x+y+z}=\dfrac{16}{16\left(x+x+y+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+2y+z}=\dfrac{16}{16\left(x+y+y+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+y+2z}=\dfrac{16}{16\left(x+y+z+z\right)}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)
Cộng từng vế của BĐT ta được :
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Vậy BĐT đã được chứng minh !
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:
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 \(x=y=z=\frac{1}{3}\)
\(xy+xz+yz=6xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{x}=a\\\frac{1}{y}=b\\\frac{1}{z}=c\end{matrix}\right.\) \(\Rightarrow a+b+c=6\)
\(T=\sum x\sqrt{\frac{x}{1+x^3}}=\sum\sqrt{\frac{x^3}{1+x^3}}=\sum\sqrt{\frac{1}{1+\frac{1}{x^3}}}=\sum\frac{1}{\sqrt{1+a^3}}=\sum\frac{1}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\Rightarrow T\ge\sum\frac{2}{a+1+a^2-a+1}=\sum\frac{2}{a^2+2}\)
Ta có đánh giá: \(\frac{2}{a^2+2}\ge\frac{7-2a}{9}\) với mọi \(0< a< 6\)
Thật vậy, \(\frac{2}{a^2+2}\ge\frac{7-2a}{9}\Leftrightarrow18-\left(a^2+2\right)\left(7-2a\right)\ge0\)
\(\Leftrightarrow2a^3-7a^2+4a+4\ge0\)
\(\Leftrightarrow\left(a-2\right)^2\left(2a+1\right)\ge0\) luôn đúng với mọi \(0< a< 6\)
Tương tự ta có: \(\frac{2}{b^2+2}\ge\frac{7-2b}{9}\) ; \(\frac{2}{c^2+2}\ge\frac{7-2c}{9}\)
\(\Rightarrow T\ge\frac{21-2\left(a+b+c\right)}{9}=\frac{21-12}{9}=1\)
\(\Rightarrow T_{min}=1\) khi \(a=b=c=2\) hay \(x=y=z=\frac{1}{2}\)