taking am-gm inequality over x times and y times b,
(a+a+a+a...upto x terms)+(b+b+b+b+....upto y terms)x+y>=[(a x a x a upto x terms)(b x b x upto y terms)]1x+y
so ax+by >= axby as x+y =1
9 Answers
since a>b.............a-b>0
hence we can divide the inequality by (a-b)
so we have to prove an-bna-b ≥n(ab)(n-1) /2
we know an-bna-b =an-1 + ban-2 + b2an-3............bn-1=k
now apply AM>gm....
am of n nos of k>their gm
try this too......
a,b,x,y all take positive rational values a±b(not equal to) and x+y=1; then prove that
ax+by >axby
one more :
if a,b ,c are sides of triangle prove that:
\prod_{cyclic}{(1+\frac{b-c}{a})^{a}} \leq 1
@xyz: for second ques..
Have you tried weighted AM-GM inequality
dunno if it will be correct or not
see this abt weighted AM-GM
the only condition is that the weights should add up to 1
http://www.artofproblemsolving.com/Wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality#Weighted_Form
that triangle one
put a=x+y , b=y+z , c=z+x , x,y,z are positive
inequality reduces to
\left(\frac{2y}{x+y} \right)^{x+y}\left(\frac{2z}{y+z} \right)^{y+z}\left(\frac{2x}{z+x} \right)}^{z+x} \le1
(y)^{x+y}(z)^{y+z}(x)^{z+x}\le\left(\frac{x+y}{2} \right)^{x+y}\left(\frac{y+z}{2} \right)^{y+z}\left(\frac{z+x}{2} \right)^{z+x} ~~~~~(1)
now from AM-GM, we have
\left(\frac{x+y}{2} \right)^{x+y} \ge (xy)^{\frac{x^y}{2}}
\left(\frac{y+z}{2} \right)^{y+z} \ge (yz)^{\frac{y+z}{2}}
\left(\frac{z+x}{2} \right)^{z+x} \ge (zx)^{\frac{z+x}{2}}
multiply the above three inequalities to get (1)
* the first of the three inequalities from AM-GM is \left(\frac{x+y}{2} \right)^{x+y} \ge (xy)^{\frac{x+y}{2}}