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Demonstrati ca: \( \frac{x}{a+x}+\frac{y}{a+y}+ \frac{z}{a+z}\leq \frac{3(x+y+z)}{3a+x+y+z}\leq \frac{x}{a+y}+\frac{y}{a+z}+\frac{z}{a+x}
\) oricare ar fi \( a,x,y,z \in (0, \infty) \)
Nela Ciceu si Titu Zvonaru

Notam m=a+x ,n=a+y ,p=a+z si prima inegalitate devine

\( \frac{m-a}{m}+\frac{n-a}{n}+\frac{p-a}{p}\le \frac{3(m+n+p-3a)}{m+n+p} \) sau

\( 3-a(\frac{1}{m}+\frac{1}{n}+\frac{1}{p})\le 3-\frac{9a}{m+n+p}\le \frac{m}{n}+\frac{n}{p}+\frac{p}{m}-a(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}) \)

Apoi prima se reduce la \( \frac{9}{m+n+p}\le\frac{1}{m}+\frac{1}{n}+\frac{1}{p} \) sau \( 9\le (m+n+p)(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}) \) sau

\( 9\le 1+1+1+(\frac{m}{n}+\frac{n}{m})+(\frac{m}{p}+\frac{p}{m})+(\frac{n}{p}+\frac{p}{n}) \) care este adevarata deoarece \( 2\le t+\frac{1}{t} \) pentru t>o.