<s>[INST] Is multiplying and dividing inequalities valid? I am facing certain problems in solving inequalities. I do not know whether multiplying and dividing inequalities works or not but it has helped me solve many problems. Here are some problems which cannot be solved using multiplying and dividing. Why does the rule not apply here?

$1$. If $a$,$b$, $c$ are positive real numbers then prove that $$(a+1)^7 (b+1)^7 (c+1)^7 > 7^7 a^4b^4c^4$$

My approach:
Applying AM-GM to $a$ and $1$ we get $(a+1) > 2\sqrt{a}$ or $(a+1)^7 > 2^7 a^\frac{7}{2}$. Proceeding similarly, we get $$(a+1)^7 (b+1)^7 (c+1)^7 > 2^{21} (abc)^\frac{7}{2}$$
Actual answer:
Apply AM-GM to $a$,$b$,$c$,$ab$,$bc$,$ca$,$abc$

$2$. If $x$ and $y$ are positive real numbers such that $x+y = 8$, then find the minimum value of $$\left(1+ \frac{1}{x}\right)\left(1+\frac{1}{y}\right)$$.

My approach:
Applying AM-GM to $1$ , $\frac{1}{x}$ and $1$ , $\frac{1}{y}$ and multiplying the inequalities we get $$\left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right) ≥ 4 \sqrt{\frac{1}{xy}}$$ Substituting $y = 8 - x$ and solving the quadratic we get $$\sqrt{\frac{1}{x(8-x)}} ≥ \frac{1}{4}$$   or the minimum value of the required expression is $1$
Actual Answer: $25/16$
 [/INST] By AM-GM $$(a+1)^7(b+1)^7(c+1)^7=\prod_{cyc}\left(4\cdot\frac{a}{4}+3\cdot\frac{1}{3}\right)^7\geq$$
$$\geq\left(7^7\right)^3\prod_{cyc}\left(\left(\frac{a}{4}\right)^4\left(\frac{1}{3}\right)^3\right)=\frac{7^{21}}{4^{12}3^9}a^4b^4c^4>7^7a^4b^4c^4.$$
The last inequality we can prove by the following way.
$$\frac{7^{21}}{4^{12}3^9}=7^7\cdot\frac{7^{14}}{2^{24}3^9}=7^7\cdot\frac{7^6\cdot7^8}{2^{24}3^9}>7^7\cdot\frac{(2^4\cdot3^8)\cdot\left(2^{11}\right)^2}{2^{24}3^9}=7^7\cdot\frac{4}{3}>7^7.$$
Now, we see why your work does not help:
we need to get $\sqrt[7]{a^4b^4c^4}$, which I got, but you got $\sqrt{a^7b^7c^7},$ which is bad.
The second inequality.
In your way you did not save the case of the equality occurring: $x=y=4$.
By your way we need $1=\frac{1}{x}$ and $1=\frac{1}{y},$ which is impossible for $x+y=8.$
Now, we can get a solution.
For $x=y=4$ we obtain a value $\frac{25}{16}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that:
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\geq\frac{25}{16}$$ or $$1+8+xy\geq\frac{25}{16}xy$$ or
$$xy\leq16,$$ which is true by AM-GM:
$$xy\leq\left(\frac{x+y}{2}\right)^2=16.$$
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