Minimum $9a+25b+49c$ when $ab+bc+ca+abc=4$ If $a,b,c\ge 0$ such that $ab+bc+ca+abc=4$, find the minimum value of $9a+25b+49c$.
I know that $a+b+c\ge 3$, but I don't this is good to use here. So I tried with Lagrange multipliers:
$$L(x,y,z)=9a+25b+49c+\lambda(ab+bc+ca+abc-4)$$
With the partial derivative I found:
$$\frac{b+c+bc}{9}=\frac{a+c+ac}{25}=\frac{a+b+ab}{49}$$
and with $ab+bc+ca+abc=4$, I found minimum $59$ at $(3,1,1/7)$. My question is, can it be done with traditional ways? I tried to prove $ab+bc+ca+abc\le 4$ when $9a+25b+49c=59$ (with idea from this question: Minimum value when $abc+ab+4bc+9ca=144$), but I got lost after expanding.
 A: An alternative solution relies on the Cauchy–Bunyakovsky–Schwarz inequality (CBS).
The given constraint $\,ab+bc+ca+abc=4\,$ can be written as
$(a+1)(b+1)(c+1) = 2+(a+1)+(b+1)+(c+1)$, which in turn is equivalent to
$$\sum_{\text{cyc}}{1\over a+2} \:=\:1\tag{1}\,.$$
Thanks to (CBS) and benefitting from $(1)$ we have
$$\begin{align*}
(3+5+7)^2 & \:=\: \left(3\sqrt{a+2}\cdot\frac1{\sqrt{a+2}}+5\sqrt{b+2}\cdot\frac1{\sqrt{b+2}}+7\sqrt{c+2}\cdot\frac1{\sqrt{c+2}}\right)^2 \\[1ex]
& \:\leqslant\: 9(a+2)+25(b+2)+49(c+2) \\[1.5ex]
\iff\quad 59 & \:\leqslant\:9a+25b+49c 
\end{align*}$$
Recall that (CBS) gives equality only if one argument is a scalar multiple of the other. This leads to the solution given in the OP.
A: A similar idea as the accepted answer after rewriting the condition as:
$$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1$$
and using Cauchy-Schwarz:
$$
\begin{aligned}
\frac{a+1}{a+2}&=\frac{1}{b+2}+\frac{1}{c+2}\\
&=5\cdot\frac{1}{5(b+2)}+7\cdot\frac{1}{7(c+2)}\\
&\geq \frac{144}{25b+49c+148}
\end{aligned}
$$
Therefore
$$25b+49c \geq \frac{144}{a+1}-4$$
and using AM-GM:
$$
\begin{aligned}
9a+25b+49c &\geq 9a+\frac{144}{a+1}-4\\
&=9(a+1)+\frac{144}{a+1}-13\\
& \geq 2\sqrt{9(a+1)\cdot \frac{144}{a+1}}-13\\
& = 59
\end{aligned}
$$
