Question

Cho các số thực không âm x, y, z sao cho x\(\le\)1, x+y\(\le\)5, x+y+z\(\le\)14. Chứng minh \(\sqrt{x}+\sqrt{y}+\sqrt{z}\le6\\ \)

Answer 1

Lời giải:
Áp dụng BĐT Bunhiacopxky:

$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq (6x+3y+2z)(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})$

Mà: $6x+3y+2z=3x+(x+y)+2(x+y+z)\leq 3.1+5+2.14=36$

Do đó: $(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq 36.(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})=36$

$\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\leq 6$ (đpcm)

Dấu "=" xảy ra khi $x=1; y=2; z=3$

Answer 2

Lời giải:
Áp dụng BĐT Bunhiacopxky:

$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq (6x+3y+2z)(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})$

Mà: $6x+3y+2z=3x+(x+y)+2(x+y+z)\leq 3.1+5+2.14=36$

Do đó: $(\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq 36.(\frac{1}{6}+\frac{1}{3}+\frac{1}{2})=36$

$\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\leq 6$ (đpcm)

Dấu "=" xảy ra khi $x=1; y=2; z=3$

Answer 3

Ý tưởng khác:

Đặt \(x=1+a;y=4+b;z=9+c\Rightarrow-1\le a\le0;b\ge-4;c\ge-9;a+b\le0;a+b+c\le0\)

C/m: \(\sqrt{1+a}+\sqrt{4+b}+\sqrt{9+c}\le6\)

\(VT=\sqrt{1\left(1+a\right)}+\frac{1}{2}\sqrt{4\left(4+b\right)}+\frac{1}{3}\sqrt{9\left(9+c\right)}\)

\(\le\frac{2+a}{2}+\frac{8+b}{4}+\frac{18+c}{6}=6+\frac{1}{12}\left(6a+3b+2c\right)\)

\(=6+\frac{1}{12}\left(3a+\left(a+b\right)+2\left(a+b+c\right)\right)\le6\)

Đẳng thức xảy ra khi \(a=b=c=0\Leftrightarrow x=1;y=4;z=9\)