3.1 

Xét hiệu :

\(\left(\dfrac{a+b}{2}\right)^2-ab=\dfrac{a^2+2ab+b^2}{4}-\dfrac{4ab}{4}\)

\(=\dfrac{a^2-2ab+b^2}{4}=\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\in R\)

Vậy \(\left(\dfrac{a+b}{2}\right)^2\ge ab,\forall a,b\in R\)

Dấu bằng xảy ra : \(\Leftrightarrow a=b\)

3.2

Áp dụng kết quả của câu 3.1 vào câu 3.2 ta được:

\(\left(a+b+c\right)^2=[a+\left(b+c\right)]^2\ge4a\left(b+c\right)\)

Mà : \(a+b+c=1\left(gt\right)\)

nên : \(1\ge4a\left(b+c\right)\)

\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\) ( vì a,b,c không âm nên b+c không âm )

Mà : \(\left(b+c\right)^2\ge4bc\Leftrightarrow\left(b-c\right)^2\ge0,\forall b,c\in N\)

\(\Rightarrow b+c\ge16abc\)

Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=c\end{matrix}\right.\Leftrightarrow b=c=\dfrac{1}{4};a=\dfrac{1}{2}\)

lần đầu tự làm được 1 bài bđt theo kiểu nháp phát đc liền... hp quớ ~~~

Đặt A = VT

từ giả thiết, ta suy ra:

\(A=\dfrac{b+c+a+b+c-2}{2+a}+\dfrac{c+a+a+b+c-3}{3+b}+\dfrac{a+b+a+b+c-4}{4+c}\)

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

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

\(=18\left(\dfrac{1}{2+a}+\dfrac{1}{3+b}+\dfrac{1}{4+c}\right)-3\)

Đặt \(B=\dfrac{1}{2+a}+\dfrac{1}{3+b}+\dfrac{1}{4+c}\)

Áp dụng bđt schwarz cho các số thực không âm:

\(B\ge\dfrac{9}{a+b+c+9}=\dfrac{1}{2}\)

vậy \(A\ge18\cdot B-3=18\cdot\dfrac{1}{2}-3=6\left(đpcm\right)\)

dấu "=" xảy ra khi \(\dfrac{1}{2+a}=\dfrac{1}{3+b}=\dfrac{1}{4+c}=\dfrac{1}{6}\) \(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\\c=2\end{matrix}\right.\)

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\)

\(\Leftrightarrow\dfrac{a^2+b^2-2ab}{ab}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{ab}\ge0\left(ab>0\right)\)

Ta có: a,b > 0

=> \(\dfrac{a}{b},\dfrac{b}{a}>0\)

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

Xét hiệu:

\(\dfrac{a}{b}+\dfrac{b}{a}-2=\dfrac{a^2+b^2-2ab}{ab}=\dfrac{\left(a-b\right)^2}{ab}\ge0\) ( luôn đúng)

\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

áp dụng BĐT cô si cho 2 số ko âm ta có

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

<=> \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) (đpcm)

Bài làm :

Ta có : \(\left(x-y\right)^2\ge0\)

\(\Rightarrow x^2+y^2\ge2xy\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow\dfrac{1}{xy}\ge\dfrac{4}{\left(x+y\right)^2}\left(1\right)\)

Áp dụng BĐT (1) ta có :

\(\dfrac{a}{b+c}+\dfrac{c}{d+a}=\dfrac{a^2+ad+bc+c^2}{\left(b+c\right)\left(d+a\right)}\ge\dfrac{4\left(a^2+ad+bc+c^2\right)}{\left(a+b+c+d\right)^2}\left(2\right)\)

Tương tự : \(\dfrac{b}{c+d}+\dfrac{d}{a+b}\ge\dfrac{4\left(b^2+ab+cd+d^2\right)}{\left(a+b+c+d\right)^2}\left(3\right)\)

Cộng các về của các BĐT (2) và (3) ta được :

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+a}+\dfrac{d}{a+b}\ge\dfrac{4\left(a^2+b^2+c^2+d^2+ad+bc+ab+cd\right)}{\left(a+b+c+d\right)^2}\)

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+a}+\dfrac{d}{a+b}\ge\dfrac{2\left(2a^2+2b^2+2c^2+2d^2+2ad+2bc+2ab+2cd\right)}{\left(a+b+c+d\right)^2}\)

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+a}+\dfrac{d}{a+b}\ge\dfrac{2[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+d\right)^2+\left(a+d\right)^2]}{\left(a+b+c+d\right)^2}=2B\)

Ta dễ dàng chứng minh được : \(B\ge1\)

Thật vậy :

\(\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+d\right)^2+\left(a+d\right)^2}{\left(a+b+c+d\right)^2}\ge1\)

\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+d\right)^2+\left(d+a\right)^2\ge\left(a+b+c+d\right)^2\)

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

\(\Rightarrowđpcm\)

Dấu đằng thức xảy ra : \(\Leftrightarrow a=c;b=d\)

khó thế tui ko hỉu

\(\Leftrightarrow\dfrac{a}{\sqrt{4b^2+bc+4c^2}}+\dfrac{b}{\sqrt{4c^2+ca+4a^2}}+\dfrac{c}{\sqrt{4a^2+ab+4b^2}}\ge1\)

Ta có:

\(\sum\left(\dfrac{a}{\sqrt{4b^2+bc+4c^2}}\right)^2\sum a\left(4b^2+bc+4c^2\right)\ge\left(a+b+c\right)^3\)

Nên ta chỉ cần chứng minh:

\(\dfrac{\left(a+b+c\right)^3}{a\left(4b^2+bc+4c^2\right)+b\left(4c^2+ac+4a^2\right)+c\left(4a^2+ab+4b^2\right)}\ge1\)

\(\Leftrightarrow\dfrac{\left(a+b+c\right)^3}{4a\left(b^2+c^2\right)+4b\left(c^2+a^2\right)+4c\left(a^2+b^2\right)+3abc}\ge1\)

\(\Leftrightarrow a^3+b^3+c^3+3abc\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\) (đúng theo Schur bậc 3)

a) theo định lý côsi :

\(\dfrac{a}{b}\)+\(\dfrac{b}{a}\)luôn >=2 với mọi a, b , a.b > 0

Ta có: (a+b)(\(\frac{1}{a}+\frac{1}{b}\ge\left(a+b\right)\frac{4}{a+b}\ge2\sqrt{\left(a+b\right)\frac{4}{\left(a+b\right)}}=2\sqrt{4}=4\)

Dấu bằng xảy ra\(\Leftrightarrow a=b;\left(a+b\right)^2=4\Leftrightarrow a=b=1\)