999 - 888 - 111 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111 + 111 - 111

= 0 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111

= 0 + 111 - 111

= 111 - 111

= 0

Bài toán sai.

Ví dụ: a \(\ge\) b \(\ge\) c  1

Thì có a=1, b=1, c=1

\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{b+1}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}<2\)

xin lỗi mk nhầm đề!!

bạn giải chi tiết ra cho mk đc ko?

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{\left(1+1+1\right)^2}{3+a+b+c+}=\frac{9}{6}=\frac{3}{2}\)

Cái đó chỉ đúng khi 1/1+a=1/1+b=1/1+c thoi

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các bạn ơi

ko bt nữa bạn nha xl nha

Áp dụng bđt Cauchy:

\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)

Tương tự:

\(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ac}{2}\)

Cộng theo vế: \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{1}{2}\left(ab+bc+ac\right)\ge3-\frac{1}{6}\left(a+b+c\right)^2=3-\frac{3}{2}=\frac{3}{2}\)\("="\Leftrightarrow a=b=c=1\)

\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=1\Rightarrow1-\dfrac{1}{1+a}=\dfrac{1}{1+b}+\dfrac{1}{1+c}\)

\(\Rightarrow\dfrac{a}{1+a}\ge\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge2\sqrt{\dfrac{1}{\left(1+b\right)\left(1+c\right)}}\) (1)

Tương tự ta có:

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

\(\dfrac{c}{1+c}\ge2\sqrt{\dfrac{1}{\left(1+a\right)\left(1+b\right)}}\) (3)

Nhân vế (1);(2);(3):

\(\Rightarrow\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Rightarrow abc\ge8\)

Dấu "=" xảy ra khi \(a=b=c=2\)

c)\(\frac{a^2}{b^2}+\frac{b^2}{a^2}+4\ge3\cdot\left(\frac{a}{b}+\frac{b}{a}\right)\)

Thế : \(\frac{\left(a-b\right)^2\left(a^2-ab+b^2\right)}{a^2b^2}\ge0\)

\(\Leftrightarrow\frac{\left(b-a\right)^2\left(a^2-ab+b^2\right)}{a^2b^2}\ge0\)

\(\Leftrightarrow\frac{a^4+4a^2b^2+b^4}{a^2b^2}\ge\frac{3\left(a^2+b^2\right)}{ab}\)

\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{a^2}+4\ge\frac{3a}{b}+\frac{3b}{a}\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{a^2}+4>=3\cdot\left(\frac{a}{b}+\frac{b}{a}\right)\)

Mấy câu khác mình đang suy nghĩ nhé

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

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

\(\Rightarrow\frac{a^2+b^2}{2}\ge\frac{\left(a+b\right)^2}{4}\)

\(\Rightarrow\frac{2\left(a^2+b^2\right)}{4}\ge\frac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(\text{a}+b\right)^2\)

Dấu ''='' chỉ xảy ra khi a=b=1 (đpcm)