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Ta có:(Sử dụng bdt cô-si) \(\frac{bc}{a^2b+a^2c}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}.\frac{b+c}{4bc}}=2.\frac{1}{2a}=\frac{1}{a}\)
=> \(\frac{bc}{a^2b+a^2c}\ge\frac{1}{a}-\frac{b+c}{4bc}\)
Chứng minh tương tự:\(\frac{ca}{b^2a+b^2c}\ge\frac{1}{b}-\frac{c+a}{4ca}\);\(\frac{ab}{c^2a+c^2b}\ge\frac{1}{c}-\frac{a+b}{4ab}\)
Từ đó \(P\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left(\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}\right)\)
Mà\(\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)=> \(P\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\ge9\)(do a+b+c<=1)=> \(P\ge\frac{1}{2}.9=\frac{9}{2}\)
Dấu '=' xảy ra <=> \(\hept{\begin{cases}a+b+c=1\\\frac{bc}{a^2b+a^2c}=\frac{b+c}{4bc}\\a,b,c>0\end{cases}};...\)
<=> \(a=b=c=\frac{1}{3}\)
Vậy\(MinP=\frac{9}{2}\)khi a=b=c=1/3
Cho $a, b, c > 0$. Đặt $P = \frac{a}{a+2b+2c} + \frac{b}{2a+b+2c} + \frac{c}{2a+2b+c}$.
Ta cần chứng minh: $\frac{3}{5} \le P < 1$.
* Chứng minh $P \ge \frac{3}{5}$:
Áp dụng bất đẳng thức Cauchy-Schwarz dạng phân thức:
$P = \frac{a^2}{a(a+2b+2c)} + \frac{b^2}{b(2a+b+2c)} + \frac{c^2}{c(2a+2b+c)}$
$\ge \frac{(a + b + c)^2}{a(a+2b+2c) + b(2a+b+2c) + c(2a+2b+c)}$
Rút gọn mẫu thức:
$a(a+2b+2c) + b(2a+b+2c) + c(2a+2b+c)$
$= a^2 + 2ab + 2ac + 2ab + b^2 + 2bc + 2ac + 2bc + c^2$
$= a^2 + b^2 + c^2 + 4ab + 4bc + 4ca$
$= (a + b + c)^2 + 2(ab + bc + ca)$
Ta có bất đẳng thức quen thuộc: $ab + bc + ca \le \frac{(a + b + c)^2}{3}$
Do đó mẫu thức:
$(a + b + c)^2 + 2(ab + bc + ca) \le (a + b + c)^2 + 2 \cdot \frac{(a + b + c)^2}{3} = \frac{5}{3}(a + b + c)^2$
Suy ra: $P \ge \frac{(a + b + c)^2}{\frac{5}{3}(a + b + c)^2} = \frac{3}{5}$
Dấu "=" xảy ra khi $a = b = c$.
Vì $a, b, c > 0$ nên:
$a + 2b + 2c > a + b + c \Rightarrow \frac{a}{a+2b+2c} < \frac{a}{a+b+c}$
$2a + b + 2c > a + b + c \Rightarrow \frac{b}{2a+b+2c} < \frac{b}{a+b+c}$
$2a + 2b + c > a + b + c \Rightarrow \frac{c}{2a+2b+c} < \frac{c}{a+b+c}$
Cộng từng vế của ba bất đẳng thức trên:
$P < \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c}$
$\Rightarrow P < \frac{a + b + c}{a + b + c} = 1$
Vậy $\frac{3}{5} \le \frac{a}{a+2b+2c} + \frac{b}{2a+b+2c} + \frac{c}{2a+2b+c} < 1$ (điều phải chứng minh).
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
\(VT=\frac{a}{a+b+a+c}+\frac{b}{a+b+b+c}+\frac{c}{a+c+b+c}\)
\(VT\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{b+c}\right)=\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)
Cho \(a=b=c\)
\(\Rightarrow2\left(\frac{a}{a+2a}+\frac{a}{a+2a}+\frac{a}{a+2a}\right)\ge1+\frac{a}{a+2a}+\frac{a}{a+2a}+\frac{a}{a+2a}\)
\(\Leftrightarrow2\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)\ge1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\)
\(\Leftrightarrow2\ge2\) ( Đúng)
\(\Rightarrow2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
Ta có:
\(\frac{a}{2a+b+c}=\frac{a}{\left(a+b\right)\left(a+c\right)}\le\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\frac{b}{a+2b+c}=\frac{b}{\left(a+b\right)\left(b+c\right)}\le\frac{b}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)
\(\frac{c}{a+b+2c}=\frac{c}{\left(a+c\right)\left(b+c\right)}\le\frac{c}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
Cộng vế theo vế:
=> \(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\le\frac{1}{4}\left(1+1+1\right)=\frac{3}{4}\)
Dấu "=" xảy ra <=> a = b = c
Cách 1:
Biến đổi tương đương bất đẳng thức cần chứng minh
\(1-\frac{a}{2b+b+c}+1-\frac{b}{a+2b+c}+1-\frac{c}{a+b+2c}\ge\frac{9}{4}\)
\(\Leftrightarrow\frac{a+b+c}{2a+b+c}+\frac{a+b+c}{a+2b+c}+\frac{a+b+c}{a+b+2c}\ge\frac{9}{4}\)
\(\Leftrightarrow4\left(a+b+c\right)\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\ge9\)
Đặt x=2a+b+c; y=a+2b+c; z=a+b+2c => x+y+z=4(a+b+c)
Khi đó đẳng thức trên trở thành
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-2\right)+\left(\frac{y}{z}+\frac{z}{y}-2\right)+\left(\frac{x}{z}+\frac{z}{x}-2\right)\ge0\)
\(\Leftrightarrow\frac{\left(x-y\right)^2}{2xy}+\frac{\left(y-z\right)^2}{2yz}+\frac{\left(z-x\right)^2}{2xz}\ge0\)
BĐT cuối luôn đúng
Vậy BĐT được chứng minh. Dấu "=" xảy ra <=> a=b=c
Cách 2:
Đặt x=2a+b+c; y=a+2b+c; z=a+b+2c
=> \(\hept{\begin{cases}a=\frac{2x-y-z}{4}\\b=\frac{3y-x-z}{4}\\c=\frac{3z-x-y}{4}\end{cases}}\)
BĐT cần chứng minh được viết lại thành
\(\frac{3x-y-z}{4x}+\frac{3y-x-z}{4y}+\frac{3z-x-z}{4z}\le\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{z}{x}\right)\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{z}{x}\ge6\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-2\right)+\left(\frac{y}{z}+\frac{z}{y}-2\right)+\left(\frac{x}{z}+\frac{z}{x}-2\right)\ge0\)
\(\Leftrightarrow\frac{\left(x-y\right)^2}{2xy}+\frac{\left(y-z\right)^2}{2yz}+\frac{\left(z-x\right)^2}{2zx}\ge0\)
BĐT cuối luôn đúng
Vậy BĐT được chứng minh. Dấu "=" <=> a=b=c
$$\sum \frac{(a-b)^2(a+b)}{2ab(b+c)} \ge 0$$Vì $a, b, c > 0$ nên bất đẳng thức trên hiển nhiên đúng.
Dấu "=" xảy ra khi $a = b = c$.
tains giải chứng mình rằng
ta có: \(\frac{b\left(2a-b\right)}{a\left(b+c\right)}=\frac{2ab-b^2}{a\left(b+c\right)}=\frac{a^2-\left(a^2-2ab+b^2\right)}{a\left(b+c\right)}=\frac{a}{b+c}-\frac{\left(a-b\right)^2}{a\left(b+c\right)}\)
ta cần CM: \(VT=\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)-\left\lbrack\frac{\left(a-b\right)^2}{a\left(b+c\right)}+\frac{\left(b-c\right)^2}{b\left(a+c\right)}+\frac{\left(c-a\right)^2}{c\left(a+b\right)}\right\rbrack\le\frac32\)
\(\sum\limits\frac{a}{b+c}-\sum\limits{\frac{\left(a-b\right)^2}{a\left(b+c\right)}}\le\frac32\) ( cậu thick tớ vt dạng tổng quát kiểu này hay bị chèn như ở trên)
<=> \(\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)-\frac32\le\frac{\left(a-b\right)^2}{a\left(b+c\right)}+\frac{\left(b-c\right)^2}{b\left(a+c\right)}+\frac{\left(c-a\right)^2}{c\left(a+b\right)}\)
ta có hằng đẳng thức
\(\sum\limits{\frac{a}{b+c}}-\frac32=\sum\limits{\frac{\left(a-b\right)^2}{2\left(b+c\right)\left(c+a\right)}}\)
CM: \(VT=\left(\frac{a}{b+c}-\frac12\right)+\left(\frac{b}{a+c}-\frac12\right)+\left(\frac{c}{a+b}-\frac12\right)\)
\(VT=\frac{2a-b-c}{2\left(b+c\right)}+\frac{2b-a-c}{2\left(a+c\right)}+\frac{2c-a-b}{2\left(a+b\right)}\)
ta có: \(2a-b-c=\left(a-b\right)+\left(a-c\right)=\left(a-b\right)-\left(c-a\right)\)
=> \(VT=\frac{\left(a-b\right)-\left(c-a\right)}{2\left(b+c\right)}+\frac{\left(b-c\right)-\left(a-b\right)}{2\left(c+a\right)}+\frac{\left(c-a\right)-\left(b-c\right)}{2\left(a+b\right)}\)
\(VT=\sum\limits{\left(a-b\right)\left\lbrack\frac{1}{2\left(b+c\right)}-\frac{1}{2\left(c+a\right)}\right\rbrack}\)
mà \(\frac{1}{2\left(b+c\right)}-\frac{1}{2\left(c+a\right)}=\frac{\left(c+a\right)-\left(b+c\right)}{2\left(b+c\right)\left(c+a\right)}=\frac{\left(a-b\right)}{2\left(b+c\right)\left(c+a\right)}\)
=> \(VT=\sum\limits{\frac{\left(a-b\right)^2}{2\left(b+c\right)\left(c+a\right)}}\)
vậy ta cần CM: \(\sum\frac{\left(a-b\right)^2}{2\left(b+c\right)\left(c+a\right)}\le\sum{\frac{\left(a-b\right)^2}{a\left(b+c\right)}}\)
<=> \(\sum\limits{\left\lbrack\frac{\left(a-b\right)^2}{a\left(b+c\right)}-\frac{\left(a-b\right)^2}{2\left(b+c\right)\left(c+a\right)}\right\rbrack\ge0}\)
\(\Leftrightarrow\sum\limits{\frac{\left(a-b\right)^2}{\left(b+c\right)}}\left\lbrack\frac{1}{a}-\frac{1}{2\left(c+a\right)}\right\rbrack\ge0\)
<=> \(\sum{\frac{\left(a-b\right)^2\left(a+2c\right)}{2a\left(b+c\right)\left(c+a\right)}}\ge0\) ( điều này luôn đúng)
dấu "=" xảy ra khi a=b=c
sau mik vt thế cho gọn:v, vt đầy đủ mỏi tay thực sự, mà mik đi về nhà nội đây:), bye cậu nha
bye nha