tau lam theo cach nay hoi dai nhung van dung

xet:a²/b²+c²-a/b+c=ab(a-b)+ac(a-c)/(b²+c²)(b+c)(1)

tg tu:b²/c²+a²-b/c+a=bc(b-c)+ab(b-a)/(a²+c²)(c+a)(2)

           c²/a²+b²-c/a+b=ac(c-a)+cb(c-b)(3)

lay(1)+(2)+(3) roi dat thua so chung ab(a-b);ac(c-a);bc(b-c) ra roi gia su a=>b=>c>0 suy ra bieu thuc trong ngoac ko am =>dpcm

Không mất tính tổng quát giả sử: \(c=min\left\{a;b;c\right\}\)chú ý rằng

\( {\displaystyle \displaystyle \sum } \)\(_{cyc}\frac{a^2+b^2}{a^2+c^2}-3=\frac{\left(a^2-b^2\right)^2}{\left(a^2+c^2\right)\left(b^2+c^2\right)}=\frac{\left(a^2-c^2\right)\left(b^2-c^2\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\)

\( {\displaystyle \displaystyle \sum }\)\(_{cyc}\frac{a+b}{b+c}-3=\frac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\frac{\left(a-c\right)\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}\)

BĐT tương đương với

\(\left(a-b\right)^2\left[\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+c\right)\left(b+c\right)}\right]+\left(a-c\right)\left(b-c\right)\)\(\left[\frac{\left(a+c\right)\left(b+c\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}-\frac{1}{\left(a+b\right)\left(b+c\right)}\right]\ge0\)

Ta có \(\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+c\right)\left(b+c\right)}\ge\frac{\left(a+b\right)^2}{\left(a+c\right)\left(b+c\right)^2}-\frac{1}{\left(a+c\right)\left(b+c\right)}\)\(=\frac{\left(a+b\right)^2-\left(a+c\right)\left(b+c\right)}{\left(a+c\right)^2\left(b+c\right)^2}\ge0\)

Ta cần chứng minh

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

\(\Leftrightarrow\frac{\left(a+c\right)^2\left(b+c\right)\left(a+b\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\ge1\)

Nếu \(a\ge b\ge c\)thì

\(\frac{\left(a+c\right)^2\left(b+c\right)\left(a+b\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\ge\frac{1}{\left(a+b\right)\left(a+c\right)}\ge\frac{\left(b+c\right)\left(a+b\right)}{a^2+b^2}\ge1\)

Nếu \(b\ge a\ge c\)thì:

\(\frac{\left(a+c\right)^2\left(b+c\right)\left(a+b\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\ge\frac{\left(b+c\right)\left(a+b\right)}{a^2+b^2}\ge\frac{b\left(a+b\right)}{a^2+b^2}\ge1\)

BĐT được chứng minh

Dấu "=" xảy ra <=> a=b=c hoặc a=b, c=0 hoặc các hoán vị tương ứng

Bạn kia chứng minh kiểu gì nhỉ, rõ ràng cho [a = 1086, b = 1000, c = 1/100] thì đề sai

Ta chứng minh BĐT sau với các số dương:

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

Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Áp dụng:

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

Cộng vế với vế:

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

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

b.

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)

\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)

Cộng vế với vế (1); (2) và (3):

\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)