Răspuns:
Explicație pas cu pas:
[tex]\displaystyle x+y\stackrel{not}{=}a,~ y+z\stackrel{not}{=}b,~ x+z\stackrel{not}{=}c\\\texttt{Inegalitatea devine:}\\\dfrac{4a+3}{b+c}+\dfrac{4b+3}{a+c}+\dfrac{4c+3}{a+b}=4\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \right)+\\3\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\geqslant 4\cdot \dfrac{3}{2}+3\cdot \dfrac{9}{2(a+b+c)}=6+\dfrac{27}{4x+4y+4z}\\\texttt{Am folosit inegalitatile}:[/tex]
[tex]\bullet \dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\geqslant \dfrac{3}{2}(\texttt{inegalitatea lui Nesbitt})\\\bullet \dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\geqslant \dfrac{(1+1+1)^2}{a+b+b+c+a+c}=\dfrac{9}{2(a+b+c)}(\texttt{aplicatie}\\\texttt{a inegalitatii lui Titu Andreescu})[/tex]
