Question

Which of the following hold good?
If sum of any two of quantities x, y, z be together greater than the third, then

• A. $(x+y+z{)}^3>27(y+z-x\left)\right(z+x-y\left)\right(x+y-z)$
  
• B. $xyz>(y+z-x)(z+x-y)(x+y-z)$
  
• C. If n is a + ive integer, then
  $(1+x^3)(1+y^3)(1+z^3)>(1+xyz{)}^3$
  
• D. $2(a^2+b^2+c^2)(a+b+c)>a^3+b^3+c^3+15abc$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(x+y+z{)}^3>27(y+z-x\left)\right(z+x-y\left)\right(x+y-z)$

B $xyz>(y+z-x)(z+x-y)(x+y-z)$

C If n is a + ive integer, then
$(1+x^3)(1+y^3)(1+z^3)>(1+xyz{)}^3$

D $2(a^2+b^2+c^2)(a+b+c)>a^3+b^3+c^3+15abc$

(a), (b), (c), (d)
(a) Now $x+y>z$ given, $\therefore x+y-z$ is +ive.
Similarly all other factors are + ive.
Put $y+z-x=a$
$z+x-y=b$
$x+y-z=c$,
$z=\frac{a+b}{2},x=\frac{b+c}{2},y=\frac{c+a}{2}$
$\therefore x+y+z=a+b+c$
Substituting in (a), we have to prove
$(a+b+c{)}^3>27abc$
But we know that $(a+b+c)>3(abc{)}^{\frac{1}{3}}$
Hence on taking cube of both sides, we get the required result.

(b) Substituting in (b), we have to prove
$\frac{a+b}{2}.\frac{b+c}{2}.\frac{c+a}{2}>abc$
which is clearly true if we apply A.M. > G.M. on each term of L.H.S. and multiply.

(c) On simplification, we have to prove that
$1+(x^3+y^3+z^3)+(x^3{y}^3+y^3{z}^3+z^3{x}^3)+x^3{y}^3{z}^3>1+3xyz+3x^2{y}^2{z}^2+x^3{y}^3{z}^3$.
Now, apply A.M. > G.M. on each bracket in L.H.S.

(d) $2(a^2+b^2+c^2)(a+b+c)>a^3+b^3+c^3-3abc+18abc$.
Now $a^3+b^3+c^3-3abc$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Transposing on the other side and taking a+b+c common, we have to prove that
$(a+b+c)(2a^2+2b^2+2c^2-a^2-b^2-c^2+ab+bc+ca)>18abc$.
Now apply A.M. > G.M. on each bracket on the left and multiply
$\therefore (a+b+c)(a^2+b^2+c^2+ab+bc+ca)>3(abc{)}^{\frac{1}{3}}.6(a^4{b}^4{c}^4{)}^{\frac{1}{6}}$
or $>18\left(abc{)}^{\frac{1}{3}}\right(abc{)}^{\frac{2}{3}}$ or $>18abc$.