Câu hỏi của Không Cần Biết 2

Question

Cho a, b, c, d > 0. Tìm giá trị nhỏ nhất của biểu thức:

$S=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{a+b+d}+\frac{d}{a+b+c}+\frac{b+c+d}{a}+\frac{c+a+d}{b}+\frac{d+a+b}{c}$

$+\frac{a+b+c}{d}$

Le Thi Khanh Huyen · Accepted Answer

Đặt $b+c+d=x;c+d+a=y;a+b+d=z;a+b+c=t$Có $a=\frac{y+z+t-2x}{3}$Tương tự :$b=\frac{x+z+t-2y}{3}$$c=\frac{x+y+t-2z}{3}$$d=\frac{y+x+z-2t}{3}$Đặt $M=\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}$Thay vào biểu thức ta có :$M=\frac{\frac{y+z+t-2x}{3}}{x}+\frac{\frac{x+z+t-2y}{3}}{y}+\frac{\frac{x+y+t-2z}{3}}{z}+\frac{\frac{y+x+z-2t}{3}}{t}$$=\frac{1}{3}\left(\frac{y+z+t-2x}{x}+\frac{x+z+t-2y}{y}+\frac{x+y+t-2z}{z}+\frac{x+z+y-2t}{t}\right)$$=\frac{1}{3}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{t}{x}+\frac{x}{t}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)+\left(\frac{t}{y}+\frac{y}{t}\right)+\left(\frac{t}{z}+\frac{z}{t}\right)-8\right]$Sử dụng BĐT Cô-si suy ra $Min_M=\frac{1}{3}.\left(12-8\right)=\frac{4}{3}$Dấu bằng xảy ra khi x = y = z = t hay $b+c+d=a+b+c=c+d+a=b+d+a$ ( tự giải ra a=b=c=d)Đặt $N=\frac{b+c+d}{a}+\frac{c+a+d}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d}$$=\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{d}{a}+\frac{a}{d}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{d}{c}+\frac{c}{d}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)$Sử dụng Cô-si ra $N\ge12$Dấu bằng xảy ra khi a=b=c=d ( tự giải ).Do đó $S=M+N\ge\frac{4}{3}+12=13\frac{1}{3}$Dấu bằng xảy ra khi $a=b=c=d$Câu trả lời: Đặt $b+c+d=x;c+d+a=y;a+b+d=z;a+b+c=t$Có $a=\frac{y+z+t-2x}{3}$Tương tự :$b=\frac{x+z+t-2y}{3}$$c=\frac{x+y+t-2z}{3}$$d=\frac{y+x+z-2t}{3}$Đặt $M=\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}$Thay vào biểu thức ta có :$M=\frac{\frac{y+z+t-2x}{3}}{x}+\frac{\frac{x+z+t-2y}{3}}{y}+\frac{\frac{x+y+t-2z}{3}}{z}+\frac{\frac{y+x+z-2t}{3}}{t}$$=\frac{1}{3}\left(\frac{y+z+t-2x}{x}+\frac{x+z+t-2y}{y}+\frac{x+y+t-2z}{z}+\frac{x+z+y-2t}{t}\right)$$=\frac{1}{3}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{t}{x}+\frac{x}{t}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)+\left(\frac{t}{y}+\frac{y}{t}\right)+\left(\frac{t}{z}+\frac{z}{t}\right)-8\right]$Sử dụng BĐT Cô-si suy ra $Min_M=\frac{1}{3}.\left(12-8\right)=\frac{4}{3}$Dấu bằng xảy ra khi x = y = z = t hay $b+c+d=a+b+c=c+d+a=b+d+a$ ( tự giải ra a=b=c=d)Đặt $N=\frac{b+c+d}{a}+\frac{c+a+d}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d}$$=\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{d}{a}+\frac{a}{d}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{d}{c}+\frac{c}{d}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)$Sử dụng Cô-si ra $N\ge12$Dấu bằng xảy ra khi a=b=c=d ( tự giải ).Do đó $S=M+N\ge\frac{4}{3}+12=13\frac{1}{3}$Dấu bằng xảy ra khi $a=b=c=d$

Kiệt Nguyễn · Answer

Áp dụng bđt cô - si cho 2 số không âm, ta được:$S=\text{ Σ}_{a,b,c,d}\left(\frac{a}{b+c+d}+\frac{b+c+d}{9a}\right)+\text{ Σ}_{a,b,c,d}\frac{8}{9}.\frac{b+c+d}{9a}$$\ge8\sqrt[8]{\frac{a}{b+c+d}.\frac{b}{c+d+a}.\frac{c}{a+b+d}.\frac{d}{a+b+c}}$$\sqrt{\frac{b+c+d}{9a}.\frac{c+d+a}{9b}.\frac{a+b+d}{9c}.\frac{a+b+c}{9d}}$$+\frac{8}{9}\left(\frac{b}{a}+\frac{c}{a}+\frac{d}{a}+\frac{c}{b}+\frac{d}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{d}{c}+\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\right)$$\ge\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}$Đẳng thức xảy ra khi a = b = c = dCâu trả lời: Áp dụng bđt cô - si cho 2 số không âm, ta được:$S=\text{ Σ}_{a,b,c,d}\left(\frac{a}{b+c+d}+\frac{b+c+d}{9a}\right)+\text{ Σ}_{a,b,c,d}\frac{8}{9}.\frac{b+c+d}{9a}$$\ge8\sqrt[8]{\frac{a}{b+c+d}.\frac{b}{c+d+a}.\frac{c}{a+b+d}.\frac{d}{a+b+c}}$$\sqrt{\frac{b+c+d}{9a}.\frac{c+d+a}{9b}.\frac{a+b+d}{9c}.\frac{a+b+c}{9d}}$$+\frac{8}{9}\left(\frac{b}{a}+\frac{c}{a}+\frac{d}{a}+\frac{c}{b}+\frac{d}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{d}{c}+\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\right)$$\ge\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}$Đẳng thức xảy ra khi a = b = c = d