Nghỉ lâu, giờ vào bài :v

Ta có : a,b,c,d >0

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

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

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

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

Cộng cả 4 vế , ta được :

\(\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}>\dfrac{a}{a+b+c+d}+\dfrac{b}{a+b+c+d}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d}=\dfrac{a+b+c+d}{a+b+c+d}=1\)Vậy \(\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}>1\left(1\right)\)

Ta lại có : \(\dfrac{a}{a+b+c}< \dfrac{a}{a+c}\)

\(\dfrac{b}{b+c+d}< \dfrac{b}{b+d}\)

\(\dfrac{c}{c+d+a}< \dfrac{c}{c+a}\)

\(\dfrac{d}{d+a+b}< \dfrac{d}{d+b}\)

Cộng 4 vế , ta được :

\(\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< \dfrac{a}{a+c}+\dfrac{b}{b+d}+\dfrac{c}{a+c}+\dfrac{d}{b+d}=\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+\left(\dfrac{b}{b+d}+\dfrac{d}{b+d}\right)=\left(\dfrac{a+c}{a+c}\right)+\left(\dfrac{b+d}{b+d}\right)=1+1=2\)

Vậy \(\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< 2\left(2\right)\)

Từ (1) và (2)=> đpcm

Bạn ơi đây là Tiếng Anh mà chứ đâu phải Toán

Áp dụng bất đẳng thức , ta có:
$VT\left[a\left(b+c\right)+b\left(c+d\right)+c\left(d+a\right)+d\left(a+b\right)\right]\ge {\left(a+b+c+d\right)}^2$
Ta cần chứng minh:
$\begin{array}{c}{\left(a+b+c+d\right)}^2\ge 2\left(ab+bc+cd+da+2ca+2bd\right)\\ \iff a^2+b^2+c^2+d^2\ge 2ca+2bd\\ \iff {\left(a-c\right)}^2+{\left(b-d\right)}^2\ge 0\end{array}$

ngu ngu ngu ngu ngu

\(\left(1-a\right)\left(1-b\right)\left(1-c\right)\left(1-d\right)\)

\(=abcd+bd+cd+ab\left(1-c\right)+ad\left(1-b\right)+ac\left(1-d\right)+bc\left(1-d\right)+\left(1-a-b-c-d\right)\)

\(>1-a-b-c-d\)

Ta có :

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)

\(\RightarrowĐPCM\)

Cho a/b= c/d (a, b, c, d khác 0) chứng minh rằng a-b/b=c-d/d

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