Dễ dàng chứng minh bất đẳng thức phụ :

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a;b>0\)và p - a; p - b; p - c > 0 theo bất đẳng thức trong tam giác.

Áp dụng bất đẳng thức phụ vừa chứng minh, ta có:

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

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

\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{2p-c-a}=\dfrac{4}{a}\left(3\right)\)

Cộng (1); (2); (3) theo vế, ta có:

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\RightarrowĐPCM\)

Ta CM BĐT sau :

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Thật vậy ; ta có :

\(\left(x-y\right)^2\ge0\\ \Rightarrow x^2-2xy+y^2\ge0\\ \Rightarrow x^2+y^2\ge2xy\\ \Rightarrow\left(x+y\right)^2\ge4xy\\ \Rightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\\ \Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\left(đpcm\right)\)

\(\Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{2p-\left(a+b\right)}=\dfrac{4}{c}\\ \dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\\ \dfrac{1}{p-a}+\dfrac{1}{p-c}\ge\dfrac{4}{b}\\ \Rightarrow2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\\ \Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(đpcm\right)\)

Tham khảo 

bạn trình bày rõ bđt 1/x + 1/y >_ 4/x+y dc ko vì mình ko hiểu lắm

\(\sqrt{\dfrac{a}{b+c-ta}}=\dfrac{a\sqrt{t+1}}{\sqrt{\left(at+a\right)\left(b+c-ta\right)}}\ge\dfrac{2a\sqrt{t+1}}{at+a+b+c-ta}=\dfrac{2a\sqrt{t+1}}{a+b+c}\)

Làm tương tự, cộng lại và rút gọn

Lời giải:

Ta có:

\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}-\frac{1}{p}=\frac{p-a+p-b}{(p-a)(p-b)}+\frac{p-(p-c)}{p(p-c)}\)

\(=\frac{2p-(a+b)}{(p-a)(p-b)}+\frac{c}{p(p-c)}=\frac{(a+b+c)-(a+b)}{(p-a)(p-b)}+\frac{c}{p(p-c)}\)

\(=\frac{c}{(p-a)(p-b)}+\frac{c}{p(p-c)}=c\left(\frac{1}{(p-a)(p-b)}+\frac{1}{p(p-c)}\right)\)

\(=c.\frac{p(p-c)+(p-a)(p-b)}{p(p-a)(p-b)(p-c)}=c.\frac{2p^2-p(a+b+c)+ab}{p(p-a)(p-b)(p-c)}\)

\(=c.\frac{2p^2-p.2p+ab}{p(p-a)(p-b)(p-c)}=\frac{abc}{p(p-a)(p-b)(p-c)}\)

Ta có đpcm.

Vì vai trò của a,b,c là như nhau, giả sử

\(a\ge c\ge b>0\)

Ta có

\(a+b-c< a\)

\(\Leftrightarrow b-c\le0\) ( đúng với gt )

\(\Rightarrow a+b-c< a\)

\(\Leftrightarrow\left(a+b-c\right)^2< a^2\)

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

CMTT :

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

Cộng vế với vế 3 BĐT trên , được

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

Ap dung bdt Cauchy-Schwarz dang Engel co:

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

Tuong tu: \(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\);

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

Cong theo ve cac bdt tren ta co:

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

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

=> Đpcm

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

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

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

@Akai Haruma