Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)Svac-so:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2\left(đpcm\right)}\)
b)\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)
\(\Leftrightarrow\dfrac{1}{a^2+1}-\dfrac{1}{ab+1}+\dfrac{1}{b^2+1}-\dfrac{1}{ab+1}\ge0\)
\(\Leftrightarrow\dfrac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{b\left(a-b\right)}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b}{\left(b^2+1\right)\left(ab+1\right)}-\dfrac{a}{\left(a^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b\left(a^2+1\right)-a\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{ab\left(a-b\right)-\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\dfrac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)(luôn đúng)
2a)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
2b)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1)
Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Giải
Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0
Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)
\(\Rightarrow\dfrac{1}{x+2y}+\dfrac{1}{2x+y}\ge\dfrac{4}{x+2y+2x+y}=\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)
Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y
Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
=> \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) = 4
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\) = 4
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2\left(\dfrac{c}{abc}+\dfrac{b}{abc}+\dfrac{a}{abc}\right)\) = 4
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}\) = 4
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2.\dfrac{abc}{abc}\) = 4 ( vì a+b + c = abc)
=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) => đpcm
Bài 1:
Áp dụng BĐt cauchy dạng phân thức:
\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\ge\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3x+3y}=4\)
dấu = xảy ra khi 2x+y=x+2y <=> x=y
Bài 2:
ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{4^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)(theo BĐt cauchy-schwarz)
\(\Rightarrow\dfrac{1}{a+b+c+d}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
Áp dụng BĐT trên vào bài toán ta có:
\(A=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)\(A\le\dfrac{1}{16}.4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
......
dấu = xảy ra khi a=b=c
Bài 2:
Áp dụng BĐT cauchy cho 2 số dương:
\(a^2+1\ge2a\)
\(\Leftrightarrow\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
thiết lập tương tự:\(\dfrac{b}{b^2+1}\le\dfrac{1}{2};\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
cả 2 vế các BĐT đều dương ,cộng vế với vế,ta có dpcm
dấu = xảy ra khi a=b=c=1
\(\text{Ta có: }\)\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Rightarrow ab+bc+ca=0\Rightarrow-ab=bc+ca\)
\(VT=\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{b^3c^3+a^3b^3+a^3c^3}{\left(abc\right)^3}\)
\(=\dfrac{\left(bc+ca\right)^3-3abc^2\left(bc+ca\right)+\left(ab\right)^3}{\left(abc\right)^3}\)
\(=\dfrac{\left(-ab\right)^3+3\left(abc\right)^2+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{\left[-\left(ab\right)^3+\left(ab\right)^3+3\left(abc\right)^2\right]}{\left(abc\right)^3}\)
\(=\dfrac{3\left(abc\right)^2}{\left(abc\right)^3}=\dfrac{3}{abc}=VP\)
Bạn tham khảo tại đây:
Câu hỏi của Hoàng Tuấn - Toán lớp 8 | Học trực tuyến
AM-GM:
\(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{b^2}\cdot\dfrac{1}{a}}=\dfrac{2}{b}\)
\(\dfrac{b}{c^2}+\dfrac{1}{b}\ge\dfrac{2}{c}\)
\(\dfrac{c}{a^2}+\dfrac{1}{c}\ge\dfrac{2}{a}\)
Cộng vế theo vế ta có:\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)
\(\Rightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)(đpcm)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{a}=x\\\dfrac{1}{b}=y\\\dfrac{1}{c}=z\end{matrix}\right.\) \(\Rightarrow x+y+z=0\) \(\Rightarrow z=-\left(x+y\right)\)
Đẳng thức cần chứng minh: \(x^3+y^3+z^3=3xyz\) với \(x+y+z=0\)
Ta có:
\(x^3+y^3+z^3=x^3+y^3-\left(x+y\right)^3=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2-\left(x+y\right)^2\right)=\left(x+y\right)\left(-3xy\right)\)
\(=-\left(x+y\right).3xy=z.3xy=3xyz\)
Vậy \(x^3+y^3+z^3=3xyz\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
\(\)
Ta có :\(\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{2}{ab}+\dfrac{2}{bc}-\dfrac{2}{ac}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2+\dfrac{2}{ab}-\dfrac{2}{bc}+\dfrac{2}{ac}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{c-a+b}{abc}\right)\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{c-\left(a-b\right)}{abc}\right)\left(1\right)\)
Theo đề ra : a=b+c
\(\Leftrightarrow c=a-b\)
\(\Leftrightarrow c-\left(a-b\right)=0\)
\(\left(1\right)\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{0}{abc}\right)=1\)
\(Hay\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1\left(đpcm\right)\)
Áp dụng bđt AM-GM: \(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{b^2a}}=2\sqrt{\dfrac{1}{b^2}}=\dfrac{2}{b}\) \(\dfrac{b}{c^2}+\dfrac{1}{b}\ge2\sqrt{\dfrac{b}{c^2b}}=2\sqrt{\dfrac{1}{c^2}}=\dfrac{2}{c}\) \(\dfrac{c}{a^2}+\dfrac{1}{c}\ge2\sqrt{\dfrac{c}{a^2c}}=2\sqrt{\dfrac{1}{a^2}}=\dfrac{2}{a}\) Cộng theo vế: \(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\Leftrightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)Dấu "=" xảy ra khi: \(a=b=c\)
Ta có :
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=1a^2+1b^2+1c^2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}\)
\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)
\(=2^2=2=2+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
\(=\dfrac{c}{abc}+\dfrac{a}{abc}+\dfrac{b}{abc}=\dfrac{abc}{abc}\)
\(=a+b+c\)
\(=abc\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\\ \Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow2+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\\ \Rightarrow\dfrac{a+b+c}{abc}=1\\ \Rightarrow a+b+c=abc\left(dpcm\right)\)