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18 tháng 2

Ta có \(VT=\dfrac{1}{a}+\dfrac{1}{4b}\)

\(=\dfrac{1}{a}+\dfrac{\dfrac{1}{4}}{b}\)

\(=\dfrac{1^2}{a}+\dfrac{\left(\dfrac{1}{2}\right)^2}{b}\)

\(\ge\dfrac{\left(1+\dfrac{1}{2}\right)^2}{a+b}\) (áp dụng BĐT \(\dfrac{x^2}{m}+\dfrac{y^2}{n}\ge\dfrac{\left(x+y\right)^2}{m+n}\))

\(=\dfrac{\left(\dfrac{3}{2}\right)^2}{1}\) (vì \(a+b=1\))

\(=\dfrac{9}{4}\)

Ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\\dfrac{1}{a}=\dfrac{1}{2b}\end{matrix}\right.\) \(\Leftrightarrow\left(a,b\right)=\left(\dfrac{2}{3},\dfrac{1}{3}\right)\)

NV
18 tháng 2

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{4b}=\dfrac{1}{a}+\dfrac{\left(\dfrac{1}{2}\right)^2}{b}\ge\dfrac{\left(1+\dfrac{1}{2}\right)^2}{a+b}=\dfrac{9}{4}\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{2}{3};\dfrac{1}{3}\right)\)

16 tháng 10 2021

help me!

AH
Akai Haruma
Giáo viên
6 tháng 1

Lời giải:

$3\text{VT}=\frac{3a}{3a+1}+\frac{3b}{3b+1}+\frac{3c}{3c+1}$

$=1-\frac{1}{3a+1}+1-\frac{1}{3b+1}+1-\frac{1}{3c+1}$

$=3-\left[\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\right]$
Áp dụng BĐT Cauchy-Schwarz:

$\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\geq \frac{9}{3a+1+3b+1+3c+1}=\frac{9}{3(a+b+c)+3}=\frac{9}{3.6+3}=\frac{3}{7}$

$\Rightarrow 3\text{VT}\leq 3-\frac{3}{7}=\frac{18}{7}$

$\Rightarrow \text{VT}\leq \frac{6}{7}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

14 tháng 12 2021

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=k\Rightarrow a=bk;b=ck;c=dk;d=ek\)

\(\Rightarrow a=bk=ck^2=dk^3=ek^4;b=ek^3\)

\(\Rightarrow\dfrac{a}{e}=\dfrac{ek^4}{e}=k^4\left(1\right)\)

Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\left(2\right)\)

Lại có \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)

19 tháng 10 2023

Ta có VP: 

\(\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)

Thay \(1=ab+bc+ca\)

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

\(=\dfrac{2}{\sqrt{\left[b\left(a+c\right)+a\left(a+c\right)\right]\left[a\left(b+c\right)+b\left(b+c\right)\right]\left[b\left(a+c\right)+c\left(a+c\right)\right]}}\)

\(=\dfrac{2}{\sqrt{\left(a+c\right)\left(a+b\right)\left(a+b\right)\left(b+c\right)\left(b+c\right)\left(a+c\right)}}\)

\(=\dfrac{2}{\sqrt{\left[\left(a+c\right)\left(a+b\right)\left(b+c\right)\right]^2}}\)

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

_____________

Ta có VT: 

\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\)

Thay \(1=ab+ac+bc\)

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

\(=\dfrac{a}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{b}{b\left(b+c\right)+a\left(b+c\right)}+\dfrac{c}{c\left(b+c\right)+a\left(b+c\right)}\)

\(=\dfrac{a}{\left(a+c\right)\left(a+b\right)}+\dfrac{b}{\left(a+b\right)\left(b+c\right)}+\dfrac{c}{\left(a+c\right)\left(b+c\right)}\)

\(=\dfrac{a\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}+\dfrac{b\left(a+c\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\dfrac{c\left(a+b\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

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

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

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

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

Mà: \(VP=VT=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\left(dpcm\right)\)

26 tháng 1 2022

:)

- Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\) (gt)

=>\(ad< bc\) 

=>\(ad+ab< bc+ab\)

=>\(a\left(b+d\right)< b\left(a+c\right)\)

=>\(\dfrac{a}{b}< \dfrac{a+c}{b+d}\) (1)

- Ta có: \(\dfrac{c}{d}>\dfrac{a}{b}\) (gt)

=>\(bc>ad\)

=>\(bc+cd>ad+cd\)

=>\(c\left(b+d\right)>d\left(a+c\right)\)

=>\(\dfrac{c}{d}>\dfrac{a+c}{b+d}\) (2)

- Từ (1) và (2) suy ra: \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

26 tháng 1 2022

- Mình lỡ làm rồi bạn tanjiro kamado gì đó :)

13 tháng 11 2021

Ta có: \(\dfrac{a}{b+c}=\dfrac{b}{a+c}\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}\left(1\right)\)

\(\dfrac{c}{a+b}=\dfrac{b}{a+c}\Rightarrow\dfrac{a+b}{c}=\dfrac{a+c}{b}\left(2\right)\)

Từ (1), (2) \(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+b}{c}=\dfrac{a+c}{b}\)