\(A=\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}\ge\dfrac{4}{2b}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{4}{b+c-a+c+a-b}\ge\dfrac{4}{2c}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{4}{a+b-c+c+a-b}\ge\dfrac{4}{2a}\ge\dfrac{2}{a}\end{matrix}\right.\)

\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Rightarrow A\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(dấu"="xảy\) \(ra\Leftrightarrow a=b=c\)

Đặt b + c - a = x; c + a - b = y; a + b - c = z. (x, y, z > 0)

Ta có \(A=\dfrac{a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{9c}{a+b-c}=\dfrac{y+z}{2x}+\dfrac{2\left(z+x\right)}{y}+\dfrac{9\left(x+y\right)}{2z}=\left(\dfrac{y}{2x}+\dfrac{2x}{y}\right)+\left(\dfrac{z}{2x}+\dfrac{9x}{2z}\right)+\left(\dfrac{9y}{2z}+\dfrac{2z}{y}\right)\ge2\sqrt{\dfrac{y}{2x}.\dfrac{2x}{y}}+2\sqrt{\dfrac{z}{2x}.\dfrac{9x}{2z}}+2\sqrt{\dfrac{9y}{2z}.\dfrac{2z}{y}}=2+3+6=11\).

Dấu "=" xảy ra khi và chỉ khi \(3y=2z=6x\Leftrightarrow3\left(c+a-b\right)=2\left(b+c-a\right)=6\left(a+b-c\right)\)

\(\Leftrightarrow a=\dfrac{5}{6};b=\dfrac{2}{3};c=\dfrac{1}{2}\).

Ta có :

\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{p-a+p-b}=\dfrac{2}{c}\)

\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{p-a+p-c}=\dfrac{2}{a}\)

\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{p-c+p-a}=\dfrac{2}{b}\)

Cộng từng về ta có đpcm

Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\left(đúng\right)\)

Áp dụng:

\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{p-a+p-b}=\dfrac{4}{2p-a-b}\)

Mà \(2p=a+b+c\)

\(\Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{a+b+c-a-b}=\dfrac{4}{c}\)

Tương tự \(\Rightarrow2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\)

\(\Rightarrowđpcm\)

bạn chứng minh :

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ( chứng minh tương tự )

ta có: \(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{p-a+p-b}=\dfrac{4}{2p-a-b}\)

mặt khác : \(p=\dfrac{a+b+c}{2}\Leftrightarrow2p=a+b+c\)

\(\Leftrightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{a+b+c-a-b}=\dfrac{4}{c}\left(1\right)\)

Chứng minh tương tự ta có:

\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\left(2\right)\)

\(\dfrac{1}{p-a}+\dfrac{1}{p-c}\ge\dfrac{4}{b}\left(3\right)\)

Cộng từng vế (1),(2),(3), ta có:

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge2\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)

\(\Leftrightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(đpcm\right)\)

BĐT cô si: \(\dfrac{x+y}{2}>\left(hoặc=\right)\sqrt{xy}\)

=>x+y >(hoặc =) \(2\sqrt{xy}\)

=>\(\left(x+y\right)^2>\left(hoặc=\right)4xy\)

=>\(\dfrac{1}{x}+\dfrac{1}{y}>\left(hoặc=\right)\dfrac{4}{x+y}\)

vì P=\(\dfrac{a+b+c}{2}=>a+b+c=2p\)

=>c=2p-a-b

b=2p-a-c

a=2p-b-c

ta có:\(\dfrac{1}{p-a}+\dfrac{1}{p-b}>hoặc=\dfrac{4}{p-a+p-b}=\dfrac{4}{c}\)

\(\dfrac{1}{p-a}+\dfrac{1}{p-c}>\left(hoặc=\right)\dfrac{4}{p-a+p-c}=\dfrac{4}{b}\)

\(\dfrac{1}{p-b}+\dfrac{1}{p-c}>\left(hoặc=\right)\dfrac{4}{p-b+p-c}=\dfrac{4}{a}\)

cộng vế với vế ta đc

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)>\left(hoặc=\right)4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

<=>\(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}>\left(hoặc=\right)2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Câu hỏi của Phạm Thị Hường - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo bài làm ở link này nhé!

Bất đẳng thức cần cm tương đương:

\(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le\dfrac{\left(a+b+c\right)^3}{27}\).

Mặt khác theo bđt AM - GM (Chú ý a, b, c là độ dài 3 cạnh của tam giác nên a + b - c > 0; b + c - a > 0; c + a - b > 0) ta có:

\(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le\dfrac{\left(a+b-c+b+c-a+c+a-b\right)^3}{27}=\dfrac{\left(a+b+c\right)^3}{27}\).

Vậy ta có đpcm.

Vì bạn không hiểu nên mình làm lại:

Thay \(p=\dfrac{a+b+c}{2}\) vào bất đẳng thức cần chứng minh ta có:

\(\left(\dfrac{a+b+c}{2}-a\right)\left(\dfrac{a+b+c}{2}-b\right)\left(\dfrac{a+b+c}{2}-c\right)\le\dfrac{\left(\dfrac{a+b+c}{2}\right)^3}{27}\)

\(\Leftrightarrow\dfrac{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}{8}\le\dfrac{\dfrac{\left(a+b+c\right)^3}{8}}{27}\)

\(\Leftrightarrow\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le\dfrac{\left(a+b+c\right)^3}{27}\).

Đến đây bạn làm tiếp như lúc nãy.