The following variables are real positive numbers, $a, b, c, d \in \mathbb{R}_{++}$ such that, the following inequality is satisfied $(ab + bc + ca)d - 3abc \leq 0$. Prove the following, $(a + b + c)d - (ab+ bc +ca) \leq 0 $.
I tried the following approach, Taking Harmonic mean of ab,bc,ca which gives $3\frac{abc}{a+b+c} \leq \frac{abc}{d}$. I obtained the following inequality from it $3d−(a+b+c) \leq 0$. After this I am stuck, do not have any idea on proceeding.
Basically you're being asked to show that $d \leq {3abc \over ab + bc + ac}$ implies $d \leq {ab + bc + ac \over a + b + c}$. This is equivalent to showing that for all positive $a,b,c$ you have $${3abc \over ab + bc + ac} \leq {ab + bc + ac \over a + b + c}$$ Let $a = {1 \over x}$, $b = {1 \over y}$, and $c = {1 \over z}$. Then what you are trying to show is that for all $x,y,z > 0$ you have $${3 \over x + y + z} \leq {x + y + z \over xy + yz + xz}$$ This can be rewritten as $$3xy + 3yz + 3xz \leq (x + y + z)^2$$ This expands to $$0 \leq x^2 + y^2 + z^2 - xy - yz - xz$$ Or equivalently $$0 \leq {1 \over 2} (x - y)^2 + {1 \over 2}(x - z)^2 + {1 \over 2}(y - z)^2$$ This trivially holds since the right-hand side is nonnegative, so we're done.