- abdellatif90 a écrit:
- soient a,b,c>=0 tq a+b+c=3
mq 1/(2+a^2+b^2)+1/(2+a^2+c^2)+1/(2+b^2+c^2)=<3/4
salam
sum( 1/(2+a²+b²) ) = sum( (a+b+c)²/(2(a+b+c)²/9 + (a²+b²))
= sum ( 9(a+b+c)² /( 2(a+b+c)² +9(a²+b²)) )
= 9 sum ( (a+b+c)²/(a(11a+2b+2c) +b(11b+2a+2c) + c(c+2a+2b) ) <= 9sum ( a²/a(11a+2b+2c) + b²/b(11b+2a+2c) + c²/c(c+2a+2b) ) ( Cauchy -shwarz) = 9* sum( a/(11a+2b+2c) + b/(11b+2a+2c) + c/(c+2a+2b) ) = 9*sum( 2a/(11a+2b+2c) + c/(c+2a+2b) )
maintenant il suffit de MQ :
sum( a/(11a+2b+2c)) <= 1/5
sum( c/(c+2a+2b) ) <= 3/5 ce qui est facile avec C.S ,je vs laisse le soin pr continuer ..
A+
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