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Đề nhảm.a;b;c ở đâu bạn -_-
a) Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel:
\(\left\{{}\begin{matrix}\dfrac{x}{2x+y+z}=\dfrac{x}{x+y+x+z}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\\\dfrac{y}{2y+x+z}=\dfrac{y}{x+y+y+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)\\\dfrac{z}{2z+x+y}=\dfrac{z}{x+z+y+z}\le\dfrac{1}{4}\left(\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\end{matrix}\right.\)
Cộng theo vế:
\(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z>0\)
b) Áp dụng bất đẳng thức AM-GM:
\(\left\{{}\begin{matrix}\left(a+b-c\right)\left(a-b+c\right)\le\dfrac{\left(a+b-c+a-b+c\right)^2}{4}=\dfrac{4a^2}{4}=a^2\\\left(a-b+c\right)\left(-a+b+c\right)\le\dfrac{\left(a-b+c-a+b+c\right)^2}{4}=\dfrac{4c^2}{4}=c^2\\\left(a+b-c\right)\left(-a+b+c\right)\le\dfrac{\left(a+b-c-a+b+c\right)^2}{4}=\dfrac{4b^2}{4}=b^2\end{matrix}\right.\)
Nhân theo vế: \(\left[\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\right]^2\le\left(abc\right)^2\)
\(\Rightarrow\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\le abc\)
Dấu "=" xảy ra khi: \(a=b=c>0\)
Phải chứng minh BĐT trung gian: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\forall\) a,b trước khi áp dụng chứ.
Ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(y+z\right)}\le\dfrac{x}{2\sqrt{\left(x+y\right)\left(y+z\right)}}\)
Tương tự với các phân số khác
\(\Rightarrow VT\le\dfrac{1}{2}\left(\dfrac{x}{\sqrt{\left(x+y\right)\left(z+x\right)}}+\dfrac{y}{\sqrt{\left(y+z\right)\left(x+y\right)}}+\dfrac{z}{\sqrt{\left(z+x\right)\left(x+y\right)}}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x+y}\cdot\sqrt{z+x}}+\dfrac{\sqrt{y}\cdot\sqrt{y}}{\sqrt{y+z}\cdot\sqrt{x+y}}+\dfrac{\sqrt{z}\cdot\sqrt{z}}{\sqrt{z+x}\cdot\sqrt{y+z}}\right)\)
\(\le\dfrac{1}{2}\left(\dfrac{\dfrac{x}{x+y}+\dfrac{x}{z+x}}{2}+\dfrac{\dfrac{y}{y+z}+\dfrac{y}{x+y}}{2}+\dfrac{\dfrac{z}{z+x}+\dfrac{z}{y+z}}{2}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{z+x}+\dfrac{x}{z+x}\right)}{2}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi x = y = z
Áp dụng tính chất dãy tỉ số bằng nhau được:
\(\dfrac{x}{2x+y+z}\)=\(\dfrac{y}{2y+x+z}\)=\(\dfrac{z}{2z+x+y}\)=\(\dfrac{x+y+z}{2x+y+z+2y+x+z+2z+x+y}\)=\(\dfrac{x+y+z}{3x+3y+3z}\)=\(\dfrac{x+y+z}{3.\left(x+y+z\right)}\)=\(\dfrac{1}{3}\)=\(\dfrac{3}{9}\)<\(\dfrac{3}{4}\)(đpcm)
Áp dụng t/c của DTSBN , ta có :
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{x+2y+z}{a+2b+c+2\left(2a+b-c\right)+4a-4b+c}\\ =\dfrac{x+2y+z}{a+2b+c+4a+2b-2a-2c+4a-4b+c}\\ =\dfrac{x+2y+z}{\left(a+4a+4a\right)+\left(2b+2b-4b\right)+\left(c-2c+c\right)}\\ =\dfrac{x+2y+z}{9a}\left(1\right)\)
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{2x+y-z}{2\left(a+2b+c\right)+2a+b-c-4a+4b+c}\\ =\dfrac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b+c}\\ =\dfrac{2x+y-z}{\left(2a+2a-4a\right)+\left(4b+b+4b\right)+\left(2c-c+c\right)}\\ =\dfrac{2x+y-z}{9b}\left(2\right)\)
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{4x-4y+z}{4\left(a+2b+c\right)-4\left(2a+b-c\right)++4a-4b+c}\\ =\dfrac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}\\ =\dfrac{4x-4y+z}{\left(4a-8a+4a\right)+\left(8b-4b-4b\right)+\left(4c+4c+c\right)}\\ =\dfrac{4x-4y+z}{9c}\left(3\right)\)
Từ (1);(2) và (3)
\(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2a+y-z}{9b}=\dfrac{4x-4y+z}{9c}\\ \Rightarrow\dfrac{x+2y+z}{9a}\cdot9=\dfrac{2a+y-z}{9b}\cdot9=\dfrac{4x-4y+z}{9c}\cdot9\\ \Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2a+y-z}{b}=\dfrac{4x-4y+z}{c}\\ \Rightarrow\dfrac{a}{a+2y+z}=\dfrac{b}{2a+y-z}=\dfrac{c}{4x-4y+z}\left(đpcm\right)\)
Đặt \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=k\left(a+2b+c\right)\\y=k\left(2a+b-c\right)\\z=k\left(4a-4b+c\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{a}{k\left(a+2b+c\right)+2k\left(2a+b-c\right)+k\left(4a-4b+c\right)}=\dfrac{a}{k.9a}=\dfrac{1}{9k}\)
Tượng tự:
\(\dfrac{b}{2x+y-z}=\dfrac{b}{9bk}=\dfrac{1}{9k}\) ; \(\dfrac{c}{4x-4y+z}=\dfrac{c}{9k.c}=\dfrac{1}{9k}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y}{2y+x+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right);\dfrac{z}{2z+y+x}\le\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)+\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}\right)=\dfrac{1}{4}\left(1+1+1\right)=\dfrac{3}{4}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{2x+y+z}=\frac{1}{(x+y)+(x+z)}\leq \frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)
\(\Rightarrow \frac{x}{2x+y+z}\leq \frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Tương tự:
\(\frac{y}{2y+x+z}\leq \frac{1}{4}\left(\frac{y}{y+z}+\frac{y}{y+x}\right)\)
\(\frac{z}{2z+x+y}\leq \frac{1}{4}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)\)
Cộng theo vế:
\(D\leq \frac{1}{4}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{4}\) (dpcm)
Dấu bằng xảy ra khi $x=y=z$
Câu b mình vừa làm rồi
a)
Áp dụng bđt Cauchy-Scharz:
\(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\)
\(=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(x+y\right)+\left(y+z\right)}+\dfrac{z}{\left(x+z\right)+\left(y+z\right)}\)
\(\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\)
\(=\dfrac{1}{4}.3=\dfrac{3}{4}\)
Dấu "=" khi \(x=y=z\)
Em ko nhớ là lớp 7 có học Cô-si nên chị đừng giải theo cách đó