another inequality from a das gupta........

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 =a^n-1 + ba^n-2 + b²a^n-3............b^n-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 >a^xb^y

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}}

anything wrong with my solution for #7 ??