Lời giải:

Từ điều kiện đề bài suy ra tồn tại các số \(x,y,z>0\) thỏa mãn:

\((a,b,c)=\left(\frac{x}{y+z},\frac{y}{x+z},\frac{z}{x+y}\right)\)

Khi đó, BĐT cần chứng minh tương đương với:

\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\geq 4\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)

\(\Leftrightarrow \left(\frac{x}{y}+\frac{x}{z}\right)+\left ( \frac{y}{x}+\frac{y}{x} \right )+\left ( \frac{z}{x}+\frac{z}{y} \right )\geq 4\left ( \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y} \right )\) \((\star)\)

BĐT trên hiển nhiên đúng do theo BĐT Cauchy-Schwarz thì:

\(\left\{\begin{matrix} x\left ( \frac{1}{y}+\frac{1}{z} \right )\geq \frac{4x}{y+z}\\ y\left ( \frac{1}{x}+\frac{1}{z} \right )\geq \frac{4y}{x+z}\\ z\left ( \frac{1}{y}+\frac{1}{x} \right )\geq \frac{4x}{y+x}\end{matrix}\right.\)

Cộng theo vế thì ta thu được \((\star)\), do đó ta có đpcm

Dấu bằng xảy ra khi \(x=y=z\Leftrightarrow a=b=c=\frac{1}{2}\)

bài h qua thì dễ mà t thì đến muộn

Bài nay khó z mak t đến sớm là sao z trời :((

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

Tưởng tìm trên mạng rồi chứ

[Toán 8] Chứng minh | Diễn đàn HOCMAI - Cộng đồng học tập lớn nhất Việt Nam

Nguyễn Huy Tú có thật đây là bài lớp 8 không

Bài 1 :

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2017}\right)\) \(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2015}{2016}.\dfrac{2016}{2017}=\dfrac{1.2.3.4....2015.2016}{2.3.4.5...2016.2017}=\dfrac{1}{2017}\)

\(B=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}....\dfrac{99^2}{99.100}\)

\(=\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}....\dfrac{99.99}{99.100}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}=\dfrac{1.2.3...99}{2.3.4...100}=\dfrac{1}{100}\)

Bài 1:a,b,c ba cạnh tam giác => a,b,c dương

\(\left\{{}\begin{matrix}a+c>b\\a+b>c\\b+c>a\end{matrix}\right.\) ta có: \(\dfrac{x}{y}< \dfrac{x+p}{y+p}\forall_{x,y,p>0\&x< y}\)

\(VT=\dfrac{a}{a+b}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+c}{a+b}+\dfrac{b}{c+a}< \dfrac{a+c+c}{a+b+c}+\dfrac{b+b}{a+b+c}=\)

\(=\dfrac{a+b+c+b+c}{a+b+c}< \dfrac{\left(a+b+c\right)+\left(A+b+c\right)}{a+b+c}< \dfrac{2\left(b+a+c\right)}{a+b+c}=2=VP\)

p/s: đề sao làm vậy:

mình nghi đề phải thế này: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\) cách làm đơn giản hơn

hướng dẫn bài 2,3 giúp mình với

Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)

Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)

\(\Rightarrowđpcm\)

Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v

Lời giải:

Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:

\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)

\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)

\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)

\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)

\(a,\\ T=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{4096}\right)\\ T=\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{4096}\right)\)

Gọi \(D=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{4096}\)

\(2D=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2048}\\ 2D-D=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2048}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{4096}\right)\\ D=1-\dfrac{1}{4096}\)

(mk nhớ có cách khác rất hay nhưng quên mất rồi)

Thay \(D\) vào ta được

\(T=\left(1+1+1+...+1\right)-\left(1-\dfrac{1}{4096}\right)\\ T=12-\left(1-\dfrac{1}{4096}\right)\\ T=12-1+\dfrac{1}{4096}\\ T=11\dfrac{1}{4096}\)

Bài 2: 

a: \(A=\dfrac{3}{2\left(x+1\right)}+\dfrac{10x}{2\left(x-1\right)\left(x+1\right)}-\dfrac{5}{2\left(x-1\right)}\)

\(=\dfrac{3x-3+10x-5x-5}{2\left(x-1\right)\left(x+1\right)}=\dfrac{8x-8}{2\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x+1}\)

b: Để P/2=3/x^2+2 thì \(\dfrac{4}{2x+2}=\dfrac{3}{x^2+2}\)

\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{3}{x^2+2}\)

=>\(2x^2+4-3x-3=0\)

=>2x^2-3x+1=0

=>(x-1)(2x-1)=0

=>x=1/2(nhận) hoặc x=1(loại)

Bài 1:

\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0

Áp dụng BĐT Chauchy cho 2 số không âm, ta có:

\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge c\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2c\)

\(\dfrac{ac}{b}+\dfrac{ab}{c}=a\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge a\sqrt{\dfrac{c}{b}.\dfrac{b}{c}}=2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}=b\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge b\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2b\)

Cộng vế theo vế ta được:

\(2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)