Exercice 1:
x,y,z trois nombres réels strictement positifs montrer que:
\(\frac{x y}{z}+\frac{y z}{y x}+\frac{z x}{y}≥x+y+z\).
* ona:
(x+z)² ≥ 0
⇾ x²+z² ≥ 2xz & y>0
⇾x²y+z²y ≥ 2xyz
⇾ x²y / xz + z²y / xz ≥ 2y
⇾ xy / z + zy / x ≥ 2y ①
de même pour
⇾ yz/x + xz/y ≥ 2z ②
⇾ zx/y + yx/z ≥ 2x ③
2xy/z + 2yz/x + 2zx/y ≥ 2x+2y+2z
Donc:
Exercice 2:
x,y,z trois nombres réels non nul
tel que: (x+y+z)² = x²+y²+z²
Montrer que:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\).
(x+y+z)²=x²+y²+z²
⇾ x²+y²+z²+2xy+2yz+2zx=x²+y²+z²
⇾ 2xy+2yz+2zx=0
⇾ \(\frac{1}{xyz}× (xy+yz+zx)= 0\) (car x,y,z ≠ 0)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
Exercice 3:
x,y,z trois nombres réels strictement positifs.
Montrer que:
\((x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})≥9\).
On a:
\((x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\)=
\(1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1\)
⇾=3+\(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\) ①
d’autre part:
on a
(x-y)²≥ 0
⇾ x²+y²≥2xy
et x,y>0
⇾ \(\frac{x^{2}}{xy}+\frac{y^{2}}{xy}\)≥2
⇾ \(\frac{x}{y}+\frac{y}{x}\)≥2
① ⇾ 3+\(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\) ≥ 3+2+2+2=9
Donc:
\((x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})≥9\)
Exercice 4:
x,y,z trois nombres réels strictement positifs
1- Montrer que:
\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}≥6\).
on a (x-y)²≥ 0
→ x²+y²≥2xy & x,y≥0
→ \(\frac{x²}{xy}+\frac{y²}{xy}≥2\)
→ \(\frac{x}{y}+\frac{y}{x}≥2\) ①
d’autre part:
on pose
A= \(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)
A= \(\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}\)
A= \((\frac{x}{z}+\frac{z}{x})+(\frac{y}{z}+\frac{z}{y})+(\frac{x}{y}+\frac{y}{x})\)
① →\(\frac{x}{z}+\frac{z}{x}≥2\) et \(\frac{y}{z}+\frac{z}{y}≥2\) et \(\frac{x}{y}+\frac{y}{x}≥2\)
Donc:
\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}≥6\)
Exercice 5:
\(x,y,z\) trois nombres réels strictement positifs et m∊IR
Tel que:
\(xyz = 1\) et \(\frac{2mx}{xy+x+1}+\frac{2my}{yz+y+1}+\frac{2mz}{zx+z+1}=1\)
Montrer que:
m= \(\frac{1}{2}\).
on pose:
\(A=\frac{2mx}{xy+x+1}+\frac{2my}{yz+y+1}+\frac{2mz}{zx+z+1}\)
On a: \(xyz = 1\)
d’ou:
\(yz = 1\)
On a: \(yz+y+1=\frac{1+xy+x}{x}\)
d’ou:
\(yz=\frac{1}{x}\)
→ \(\frac{2my}{yz+y+1}=\frac{2my}{\frac{1}{x}+y+1}\)
→ \(\frac{2my}{yz+y+1}=\frac{2mxy}{1+xy+x}\) ①
d’autre part
\(\frac{2mz}{zx+z+1}=\frac{2mz}{z(x+1+\frac{1}{z})}\)
\(=\frac{2m}{(x+1+\frac{1}{z}}\)
\(\frac{2mz}{zx+z+1}=\frac{2m}{x+1+xy}\)②
(car \(xy=\frac{1}{z}\) )
Alors:
puis que A=1
Alors
\(A=\frac{2mx}{xy+x+1}+\frac{2mxy}{1+xy+x}+\frac{2m}{x+1+xy}\)
\(=\frac{2mx+2mxy+2m}{xy+x+1}\)
\(=2m \frac{x+xy+1}{xy+x+1}\)
\(=2m =1\)
Donc:
\(m= \frac{1}{2}\)