Réponse :

Bonjour,

Explications étape par étape :

1)

[tex]x,y > 0 \\\\(x-y)^2 \geq 0\\\\x^2+y^2-2xy \geq 0\\\\x^2+y^2 \geq 2xy\\\\\dfrac{x^2+y^2}{xy} \geq 2\\\\\boxed{\dfrac{x}{y}+ \dfrac{y}{x} \geq 2 }\\[/tex]

2)

[tex]\dfrac{a+b}{a+c} +\dfrac{a+c}{a+b} \geq 2\\\\\dfrac{a+b}{b+c} +\dfrac{b+c}{a+b} \geq 2\\\\\dfrac{a+c}{b+c} +\dfrac{b+c}{a+c} \geq 2\\\\\\(\dfrac{a+b}{a+c} +\dfrac{b+c}{a+c} )+ (\dfrac{a+c}{a+b} +\dfrac{b+c}{a+b} )+(\dfrac{a+b}{b+c} +\dfrac{a+c}{b+c} ) \geq 6\\\\\\\dfrac{a+2b+c}{a+c} +\dfrac{a+b+2c}{a+b} + \dfrac{2a+b+c}{b+c} \geq 6\\\\[/tex]

[tex]2\dfrac{b}{a+c}+1 +2 \dfrac{c}{a+b} +1+ 2\dfrac{a}{b+c} +1 \geq 6\\\\2(\dfrac{b}{a+c} +\dfrac{c}{a+b} + \dfrac{a}{b+c} )+3 \geq 6\\\\2(\dfrac{b}{a+c} +\dfrac{c}{a+b} + \dfrac{a}{b+c} ) \geq 3\\\\\dfrac{b}{a+c} +\dfrac{c}{a+b} + \dfrac{a}{b+c} \geq \dfrac{3}{2}\\[/tex]