Bài làm:

Ta có: \(S=\frac{a-d}{b+d}+\frac{d-b}{c+b}+\frac{b-c}{a+c}+\frac{c-a}{d+a}\)

\(S=\left(\frac{a-d}{b+d}+1\right)+\left(\frac{d-b}{c+b}+1\right)+\left(\frac{b-c}{a+c}+1\right)+\left(\frac{c-a}{d+a}+1\right)-4\)

\(S=\frac{a+b}{b+d}+\frac{c+d}{c+b}+\frac{a+b}{a+c}+\frac{c+d}{d+a}-4\)

\(S=\left(a+b\right)\left(\frac{1}{b+d}+\frac{1}{a+c}\right)+\left(c+d\right)\left(\frac{1}{c+b}+\frac{1}{d+a}\right)-4\)

\(\ge\left(a+b\right)\frac{\left(1+1\right)^2}{a+b+c+d}+\left(c+d\right)\frac{\left(1+1\right)^2}{a+b+c+d}-4\)

\(=\frac{4\left(a+b\right)}{a+b+c+d}+\frac{4\left(c+d\right)}{a+b+c+d}-4=\frac{4\left(a+b+c+d\right)}{a+b+c+d}-4=4-4=0\)

Dấu "=" xảy ra khi: \(a=b=c=d\)

Vậy \(Min\left(S\right)=0\Leftrightarrow a=b=c=d\)

Học tốt!!!!

Để \(\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}\ge\frac{a-d}{a+b}\)

\(\Leftrightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge0\)

\(\Leftrightarrow\frac{a-b}{b+c}+1+\frac{b-c}{c+d}+1+\frac{c-d}{d+a}+1+\frac{d-a}{a+b}+1\ge4\)

\(\Leftrightarrow\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\ge4\)

\(\Leftrightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge4\)(Cần phải chứng minh)

Ta có : \(\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\ge\left(a+c\right).\frac{4}{a+b+c+d}\left(1\right)\)(Áp dụng BĐT Cô-si)

\(\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge\left(b+d\right).\frac{4}{a+b+c+d}\left(2\right)\)(Áp dụng BĐT Cô-si)

Từ (1) và (2) \(\Rightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\)

\(\ge\frac{4\left(a+c\right)}{a+b+c+d}+\frac{4\left(b+d\right)}{a+b+c+d}=4\)(Điều phải chứng minh)

Thank bạn Fire Sky very much ☺☺🙂☺☺!!

đpcm<=>(\(\frac{a}{b+c+d}\)-\(\frac{1}{3}\))+(\(\frac{b}{a+c+d}\)-\(\frac{1}{3}\))+(\(\frac{c}{a+b+d}\)-\(\frac{1}{3}\))+(\(\frac{d}{a+b+c}\)-\(\frac{1}{3}\))\(\ge\)0

Xét giá trị của các dấu ngoặc,dễ thấy chúng đều lớn hơn hoặc bằng 0

Vậy thì bất đẳng thức trên là đúng hay đpcm là đúng

khoannnnnnnn, bn: Lê Hồ Trọng Tín ơi:

nếu a=1,b=2,c=1,d=1 thì: \(\frac{1}{2+1+1}=\frac{1}{4}-\frac{1}{3}\ge0???\)

mọe, t-i-k đúng nhầm :(((

Đặt \(\hept{\begin{cases}b+c-a=2x\\c+a-b=2y\\a+b-c=2z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=y+z\\b=x+z\\c=x+y\end{cases}}\)

\(\Rightarrow P=\frac{1}{2}.\left(\frac{4\left(y+z\right)}{x}+\frac{9\left(x+z\right)}{y}+\frac{16\left(x+y\right)}{z}\right)\)

\(=\frac{1}{2}.\left(\left(\frac{4y}{x}+\frac{9x}{y}\right)+\left(\frac{4z}{x}+\frac{16x}{z}\right)+\left(\frac{9z}{y}+\frac{16y}{z}\right)\right)\)

\(\ge\frac{1}{2}.\left(2.2.3+2.2.4+2.3.4\right)=26\)

nỏ biết

Lời giải:

a)

\(A=4x^2-4x+1=2x(2x-3)+2x+1=2x(2x-3)+(2x-3)+4\)

\(=(2x+1)(2x-3)+4\)

Với \(x\geq \frac{3}{2}\Rightarrow \left\{\begin{matrix} 2x+1>0\\ 2x-3\geq 0\end{matrix}\right.\Rightarrow A=(2x+1)(2x-3)+4\geq 4\)

Vậy GTNN của $A$ là $4$ khi $x=\frac{3}{2}$

b)

\(B=5x^2-10x+3=5(x^2-2x+1)-2\)

\(=5(x-1)^2-2\)

Ta thấy \((x-1)^2\geq 0, \forall x\geq 1\Rightarrow B=5(x-1)^2-2\geq -2\)

Vậy GTNN của $B$ là $-2$ khi $(x-1)^2=0\Leftrightarrow x=1$

c)

\(C=4x^2-6x+2=(2x)^2-2.2x.\frac{3}{2}+(\frac{3}{2})^2-\frac{1}{4}\)

\(=(2x-\frac{3}{2})^2-\frac{1}{4}\)

Ta thấy \((2x-\frac{3}{2})^2\geq 0, \forall x\geq 0\Rightarrow C=(2x-\frac{3}{2})^2-\frac{1}{4}\geq -\frac{1}{4}\)

Vậy GTNN của $C$ là $\frac{-1}{4}$ khi \((2x-\frac{3}{2})^2=0\Leftrightarrow x=\frac{3}{4}\)

d)

\(D=3x^2+2x+1=3(x^2+\frac{2}{3}x+\frac{1}{9})+\frac{2}{3}\)

\(=3(x+\frac{1}{3})^2+\frac{2}{3}\)

Ta thấy \((x+\frac{1}{3})^2\geq 0, \forall x\geq -1\Rightarrow D=3(x+\frac{1}{3})^2+\frac{2}{3}\geq \frac{2}{3}\)

Vậy GTNN của $D$ là $\frac{2}{3}$ khi $(x+\frac{1}{3})^2=0\Leftrightarrow x=-\frac{1}{3}$

Không mất tính tổng quát, giả sử \(a+b\ge c+d\)

Từ giả thiết suy ra \(b+c\ge\frac{a+b+c+d}{2}\)

\(A=\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-\left(\frac{c}{c+d}-\frac{c}{a+b}\right)\)

\(\ge\frac{a+b+c+d}{2\left(c+d\right)}-\left(\frac{c+d}{c+d}-\frac{c+d}{a+b}\right)\)

Đặt a + b = x ; c + d = y ( \(x\ge y>0\), ta có :

\(A\ge\frac{x+y}{2y}-\frac{y}{y}+\frac{y}{x}=\frac{x}{2y}+\frac{1}{2}-1+\frac{y}{x}=\left(\frac{x}{2y}+\frac{y}{x}\right)-\frac{1}{2}\ge2\sqrt{\frac{x}{2y}.\frac{y}{x}}-\frac{1}{2}=\sqrt{2}-\frac{1}{2}\)

Vậy GTNN của A là \(\sqrt{2}-\frac{1}{2}\Leftrightarrow d=0,x=y\sqrt{2};b+c=a+d\)

chẳng hạn \(a=\sqrt{2}+1;b=\sqrt{2}-1;c=2;d=0\)

Ta có

\(4\left(a+b+c+d\right)^2=\left(\left(a+b\right)+\left(b+c\right)+\left(c+d\right)+\left(d+a\right)\right)^2\)

\(=\left(\frac{\sqrt{a+b}}{\sqrt{b+c+d}}.\sqrt{a+b}.\sqrt{b+c+d}+\frac{\sqrt{b+c}}{\sqrt{c+d+a}}.\sqrt{b+c}.\sqrt{c+d+a}+\frac{\sqrt{c+d}}{\sqrt{d+a+b}}.\sqrt{c+d}.\sqrt{d+a+b}+\frac{\sqrt{d+a}}{\sqrt{a+b+c}}.\sqrt{d+a}.\sqrt{a+b+c}\right)^2\)

\(\le\left(\frac{a+b}{b+c+d}+\frac{b+c}{c+d+a}+\frac{c+d}{d+a+b}+\frac{d+a}{a+b+c}\right)\left(\left(a+b\right)\left(b+c+d\right)+\left(b+c\right)\left(c+d+a\right)+\left(c+d\right)\left(d+a+b\right)+\left(d+a\right)\left(a+b+c\right)\right)\)

\(\Rightarrow VT\ge\frac{4\left(a+b+c+d\right)^2}{\left(\left(a+b\right)\left(b+c+d\right)+\left(b+c\right)\left(c+d+a\right)+\left(c+d\right)\left(d+a+b\right)+\left(d+a\right)\left(a+b+c\right)\right)}\)(1)

Ta chứng minh

\(4\left(a+b+c+d\right)^2\ge\frac{8}{3}\left(\left(a+b\right)\left(b+c+d\right)+\left(b+c\right)\left(c+d+a\right)+\left(c+d\right)\left(d+a+b\right)+\left(d+a\right)\left(a+b+c\right)\right)\left(2\right)\)

\(\Leftrightarrow a^2+b^2+c^2+d^2-2ac-2bd\ge0\)

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

Từ (1) và (2) ta

\(\Rightarrow\frac{a+b}{b+c+d}+\frac{b+c}{c+d+a}+\frac{c+d}{d+a+b}+\frac{d+a}{a+b+c}\ge\frac{8}{3}\)

Dấu = xảy ra khi a = b = c = d

dau = xay ra khi a=b=c=d

de qua tu tinh len mang ma tra tao day ko muon giai