4 replies to this topic

[lahantaithe99]

Cho $x,y,z>0$ thỏa mãn $x+y+z=xyz$

Tìm Max  $\frac{2}{\sqrt{x^2+1}}+\frac{1}{\sqrt{y^2+1}}+\frac{1}{\sqrt{z^2+1}}$

[nguyenqn1998]

Cho $x,y,z>0$ thỏa mãn $x+y+z=xyz$

Tìm Max  $\frac{2}{\sqrt{x^2+1}}+\frac{1}{\sqrt{y^2+1}}+\frac{1}{\sqrt{z^2+1}}$

Do $x+y+z=xyz$ nên tồn tại $tanA,tanB,tanC$ sao cho:

$x=tanA,y=tanB,z=tanC$

$=> \frac{2}{\sqrt{x^2+1}}+\frac{1}{\sqrt{y^2+1}}+\frac{1}{\sqrt{z^2+1}}=2.cosA+cosB+cosC=2.(1-2sin^2\frac{A}{2})+2cos(\frac{B+C}{2}).cos(\frac{B-C}{2})\leq 2-4sin^2\frac{A}{2}+2.sin\frac{A}{2}=2,5-(2sin\frac{A}{2}-\frac{1}{2})^2\leq 2,5$

Edited by nguyenqn1998, 19-01-2014 - 09:51.

[lahantaithe99]

Do $x+y+z=xyz$ nên tồn tại $tanA,tanB,tanC$ sao cho:

$x=tanA,y=tanB,z=tanC$

$=> \frac{2}{\sqrt{x^2+1}}+\frac{1}{\sqrt{y^2+1}}+\frac{1}{\sqrt{z^2+1}}=2.cosA+cosB+cosC=2.(1-2sin^2\frac{A}{2})+2cos(\frac{B+C}{2}).cos(\frac{B-C}{2})\leq 2-4sin^2\frac{A}{2}+2.sin\frac{A}{2}=2,5-(2sin\frac{A}{2}-\frac{1}{2})^2\leq 2,5$

Dấu = xảy ra khi nào ạ?

Đây là bài trong box THCS cách làm của anh em vẫn chưa hiểu

[qwerty98]

$x+y+z=xyz =>x^{2}+xy+xz=x^{2}yz=>yz(x^{2}+1)=(x+z)(x+y) A=\sqrt{\frac{4yz}{(x+y)(x+z)}}+\sqrt{\frac{2xz}{(x+y)2(y+z)}}+\sqrt{\frac{2xy}{(z+x)2(z+y)}}\leq \frac{y}{x+y}+\frac{z}{x+z}+\frac{x}{x+y}+\frac{z}{4(y+z)}+\frac{x}{x+z}+\frac{y}{4(y+z)}=9/4$

Edited by qwerty98, 19-01-2014 - 21:32.

[nguyentrungphuc26041999]

Cho $x,y,z>0$ thỏa mãn $x+y+z=xyz$

Tìm Max  $\frac{2}{\sqrt{x^2+1}}+\frac{1}{\sqrt{y^2+1}}+\frac{1}{\sqrt{z^2+1}}$

Đặt $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$

ta có $ab+bc+ca=1$

trên đề 

$\frac{2}{\sqrt{\frac{1}{a^{2}}+1}}+\frac{1}{\sqrt{\frac{1}{b^{2}}+1}}+\frac{1}{\sqrt{\frac{1}{c^{2}}+1}}=\frac{2a}{\sqrt{a^{2}+ab+bc+ca}}+\frac{b}{\sqrt{b^{2}+ab+bc+ca}}+\frac{c}{\sqrt{c^{2}+ab+bc+ca}}$

$\frac{2a}{\sqrt{\left ( a+b \right )\left ( a+c \right )}}+\frac{2b}{\sqrt{4\left ( b+c \right )\left ( b+a \right )}}+\frac{2c}{\sqrt{4\left ( c+a \right )\left ( c+b \right )}}\leq \frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+a}+\frac{b}{4\left ( b+c \right )}+\frac{c}{a+c}+\frac{c}{4\left ( b+c \right )}=\frac{9}{4}$