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?

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

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

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

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)

Với a,b,c > 0 áp dụng BĐT Cauchy, ta có

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)

Cmtt: \(\dfrac{c}{a}+\dfrac{a}{c}\ge2\) và \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\)

Theo đề bài, ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)(do a + b + c = 1)

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

\(=3+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}\)\(\ge3+2+2+2=9\)

\(a\left(a-b\right)+b\left(b-c\right)+c\left(c-a\right)\ge0\)

\(\Leftrightarrow a^2-ab+b^2-bc+c^2-ac\ge0\)

\(\Leftrightarrow2a^2-2ab+2b^2-2bc+2c^2-2ac\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\)

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

Vậy \(a\left(a-b\right)+b\left(b-c\right)+c\left(c-a\right)\ge0\)

khó quá

Áp dụng bất đẳng thức Cauchy - Schwarz dưới dạng Engel ta có :

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

Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)

ko biét

Ta có:

\(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)+a^2+b^2+c^2=\left(a+b+c\right)^2\)

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