39is it for positive reals or is it for all real numbers x,y,z ??
please be specific with such conditions while giving an inequality problem...
this simplifies the solution a lot in most of the cases.
prove that
xyz >(y+z-x)(z+x-y)(x+y-z)
i have found one solution but i am sure a better is existing in someones brain :)
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9 Answers
39
1Considering x,y,z as positive reals,
taking AM -GM on x+z-y and y+z-x ,
we get y+z-x+z+x-y2 ≥ √(z+x-y)(y+z-x)
or , z ≥ √(z+x-y)(y+z-x)
similarly by taking x+z-y and x+y-z , x ≥ √(x+y-z)(x+z-y)
and by y+z-x and x+y-z , y ≥ √(y+z-x)(x+y-z)
multiplying all these , xyz ≥ (y+z-x)(z+x-y)(x+y-z)
1no problem at all , that is why we are here , to help each other out :):):):):)
1well here is my solution
we need t prove :
1\geq (\frac{y}{x}+\frac{z}{x}-1)(\frac{z}{y}+\frac{x}{y}-1)(\frac{x}{z}+\frac{y}{z}-1)
expanding the brackes and using the symmetrical arguments.,the expansion reduces to
1\geq \sum{\left(cyclic \right)(\frac{z}{x}+\frac{x}{z})}-\sum{\left(cyclic \right)\frac{y^{2}}{zx}} -2
this is nothing but twice of cosine formula
1 \geq 2(cosA +cosB +cosC)-2
where A,B,C are sides of a triangle
which is obvious
3well can we have x+z>y for all nos and we need a proof for all real nos.Both of u assumed the same dat x+z>y,i guess its not the exact proof.
1soumya i dont think we can apply AM-GM to x+y-z , ... and so on as you dont know all the terms are positive. another soln
put x=a+b ,\ y=b+c,\ z=a+c for some positive reals a,b,c
the inequality reduces to
\left(\frac{a+b}{2} \right)\left(\frac{b+c}{2} \right)\left(\frac{a+c}{2} \right) \ge abc
which is true by AM-GM on individual brackets
62@xyz.. i think there is somethign wrong.. either with the conditions you have mentioned.. or something...
as all the guys above have pointed..
for the most simple cases.. try x=0
or x=y=z
The inequality does not hold..
