maximize $M+N+P$ with the restraint $5(MN+MP+PN)=7MNP$ and $M,N,P$ are positive integers Mal, Num, and Pin each have distinct number of marbles. Five times the sum of
the product of the number of marbles of any two of them equals to seven times
the product of the number of the marbles the three of them have. Find the largest
possible sum of their marbles. 

the question is essentially maximize $M+N+P$ with the restraint $5(MN+MP+PN)=7MNP$ and $M,N,P$ are positive integers

I tried using lagrange multipliers but it didn't seem to work, I think because the degree of the expression is 1.

the  expression shows that
atleast one of the variables are a multiple of 5 

hints, suggestions and solutions would all be appreciated

taken from the 2017 InImc https://chiuchang.org/wp-content/uploads/sites/2/2018/01/2017_IWYMIC_answer-2.x17381.pdf
Actually, we can do this question in a simple way.
WLOG, let $M&lt;N&lt;P$.

$\dfrac{1}{M}+\dfrac{1}{N}+\dfrac{1}{P}=\dfrac{MN+MP+NP}{MNP}=\dfrac{7}{5}$

If $M=2$, then $\dfrac{1}{M}+\dfrac{1}{N}+\dfrac{1}{P}&lt;\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{13}{12}&lt;\dfrac{7}{5}$, therefore $M=1, \dfrac{1}{N}+\dfrac{1}{P}=\dfrac{2}{5}$

Rearrange it and we get $\left(2N-5\right)\left(2P-5\right)=25$

$\because N&lt;P \\ \therefore \begin{cases} 2N-5=1 \\ 2P-5=25 \end{cases} \rightarrow \begin{cases} N=3 \\ P=15 \end{cases}$

$M+N+P=1+3+15=\boxed{19}$
Was the constraint $5(MN+ MP+ PN)= 7MNP$?  What you have "$5(MN+ MP+ PM)= 7MNP$"  doesn't make sense (MP+ PM= 2MP).. Using Lagrange multipliers write $f(M,N,P)= M+ N+ P$  and $g(M,N,P)= 5(MN+ MP+ PN)- 7MNP$. Then $\nabla f= &lt;1, 1, 1&gt;$ and $\nabla g= &lt;5N+ 5P- 7NP, 5M+ 5P- 7MP, 5M+ 5N- 7MN&gt;$. So we must have $5N+ 5P- 7NP= \lambda$, $5M+ 5P- 7MP= \lambda$, and $5M+ 5N- 7MN=\lambda$.  Since those are all equal to $\lambda$, they are all equal to each other:
5N+ 5P- 7NP= 5M+ 5P- 7MP so 5N- 7NP= 5M- 7MP or 5(N-M)= 7P(N-M) or, as long as N is not equal to M, 5= 7P and P= 5/7.  And 5M+ 5P- 7MP= 5M+ 5N- 7MN so 5P- 7MP= 5N- 7MN or 5(P- M)= 7M(P- M) so, as long as P is not equal to M, M= 5/7.  Also, of course, M, N, and P must satisfy the restraint 5MN+ 5MP+ 5PN= 7MNP.  We also have the constraint, 5(MN+ MP+ PN)= 7MNP.