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

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+3\ge\dfrac{2\left(a+b+c\right)}{abc}=2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xyz=1\)

BĐT trở thành: \(x^2+y^2+z^2+3\ge2\left(xy+yz+zx\right)\)

Theo nguyên lý Dirichlet, trong 3 số x;y;z luôn có ít nhất 2 số cùng phía so với 1

Không mất tính tổng quát, giả sử đó là x và y \(\Rightarrow\left(x-1\right)\left(y-1\right)\ge0\)

\(\Rightarrow xy+1\ge x+y\Rightarrow xyz+z\ge xz+yz\Rightarrow2xyz+2z\ge2xz+2yz\)

\(\Rightarrow2\ge2xz+2yz-2z\) (do \(xyz=1\))

\(\Rightarrow VP=x^2+y^2+z^2+2+1\ge x^2+y^2+z^2+2xz+2yz-2z+1\)

\(VP\ge2xy+z^2+2xz+2yz-2z+1=2\left(xy+yz+zx\right)+\left(z-1\right)^2\ge2\left(xy+yz+zx\right)\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

Mình làm được rồi, cảm ơn các bạn

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

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)

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

\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)

Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)

Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

\(\dfrac{a}{bc}+\dfrac{b}{ac}>=2\cdot\sqrt{\dfrac{a}{bc}\cdot\dfrac{b}{ac}}=\dfrac{2}{cc}\)

\(\dfrac{b}{ca}+\dfrac{c}{ab}>=2\cdot\sqrt{\dfrac{bc}{ca\cdot ab}}=\dfrac{2}{a}\)

\(\dfrac{c}{ab}+\dfrac{a}{bc}>=2\cdot\sqrt{\dfrac{a\cdot c}{a\cdot b\cdot c\cdot b}}=\dfrac{2}{b}\)

=>a/bc+b/ac+c/ab>=2(1/a+1/b+1/c)

Ta có: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\sqrt{\dfrac{2c}{a+b}}\)

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

\(\ge\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b+2c}=\dfrac{\left(a-b\right)^2\left(a+b+c\right)}{\left(b+c\right)\left(c+a\right)\left(a+b+2c\right)}\ge0\)

(đúng hiển nhiên)

Đẳng thức xảy ra khi $a=b=c.$

\(\left(1+\dfrac{1}{n}\right)^n=C_n^0+C_n^1.\dfrac{1}{n}+C_n^2.\dfrac{1}{n^2}+...+C_n^n.\dfrac{1}{n^n}\)

\(=1+1+C_n^2.\dfrac{1}{n^2}+C_n^3.\dfrac{1}{n^3}+...+C_n^n.\dfrac{1}{n^n}\)

\(=2+C_n^2.\dfrac{1}{n^2}+C_n^3.\dfrac{1}{n^3}+...+C_n^n.\dfrac{1}{n^n}>2\)

Mặt khác:

\(C_n^k.\dfrac{1}{n^k}=\dfrac{n!}{k!\left(n-k\right)!.n^k}=\dfrac{\left(n-k+1\right)\left(n-k+2\right)...n}{n^k}.\dfrac{1}{k!}< \dfrac{n.n...n}{n^k}.\dfrac{1}{k!}=\dfrac{n^k}{n^k}.\dfrac{1}{k!}=\dfrac{1}{k!}\)

\(< \dfrac{1}{k\left(k-1\right)}=\dfrac{1}{k-1}-\dfrac{1}{k}\)

Do đó:

\(C_n^2.\dfrac{1}{n^2}+C_n^3.\dfrac{1}{n^3}+...+C_n^n.\dfrac{1}{n^n}< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)

\(\Rightarrow2+C_n^2.\dfrac{1}{n^2}+C_n^3.\dfrac{1}{n^3}+...+C_n^n.\dfrac{1}{n^n}< 2+1=3\) (đpcm)