Réponse :
Bonsoir,
Explications étape par étape :
1)
Comme x est strictement positif:
[tex](x-1)^2\geq 0\\\Longrightarrow\ x^2-2x+1\geq 0\\\Longrightarrow\ x^2+1\geq 2x\\\Longrightarrow\ \dfrac{x^2+1}{x} \geq 2\\\Longrightarrow\ x+\dfrac{1}{x} \geq 2\\[/tex]
2)
[tex](x+\dfrac{1}{x} )+(y+\dfrac{1}{y} )+(z+\dfrac{1}{z} ) \geq 2+2+2\\\\x+y+z+\dfrac{1}{x} +\dfrac{1}{y} +\dfrac{1}{z} \geq 6\\\\\\(x+y+z)*(\dfrac{1}{x} +\dfrac{1}{y} +\dfrac{1}{z} )\\=1+\dfrac{y}{x} +\dfrac{z}{x} +\dfrac{x}{y}+1 +\dfrac{z}{y} +\dfrac{x}{z} +\dfrac{y}{z} +1\\=3+(\dfrac{x}{y} +\dfrac{1}{\dfrac{x}{y}})+(\dfrac{x}{z} +\dfrac{1}{\dfrac{z}{x}})+(\dfrac{y}{z} +\dfrac{1}{\dfrac{z}{y}})\geq 3+2+2+2\\\\(x+y+z)*(\dfrac{1}{x} +\dfrac{1}{y} +\dfrac{1}{z} )\geq 6[/tex]
3)
[tex](x-y)^2\geq 0\\\Longrightarrow\ x^2+y^2-2xy\geq 0\\\Longrightarrow\ 2xy\leq x^2+y^2\\\Longrightarrow\ 4xy\leq x^2+y^2+2xy\\\Longrightarrow\ 4xy\leq (x+y)^2\\\\\Longrightarrow\ \dfrac{xy}{x+y} \leq \dfrac{x+y}{4}[/tex]
4)
[tex]\dfrac{xy}{x+y} + \dfrac{yz}{y+z}+\dfrac{xz}{x+z} \leq \dfrac{x+y}{4} +\dfrac{y+z}{4} +\dfrac{x+z}{4} \\\\\\\dfrac{xy}{x+y} + \dfrac{yz}{y+z}+\dfrac{xz}{x+z} \leq \dfrac{x+y+z}{2}[/tex]
