[Inequalities] given a,b,c are distinct positive integers

[rain berry]

so tatsu hasnt been disabled here

,tex test

oh wait we knew texit worked

[eternal imp]

anyone to say anything now is gay

[coarse anvil]

[valid sandal]

We can start by expanding the right-hand side of the given inequality to get:

$$\frac{3}{2} + \frac{a+b+c}{a+b+c+4abc} = \frac{3a+3b+3c+a+b+c+4abc}{2a+2b+2c+4abc}$$

Now, we can use the fact that $a$, $b$, and $c$ are positive integers to write the numerator and denominator of the fraction on the left-hand side of the given inequality in a similar form:

$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = \frac{a(c+a) + b(a+b) + c(b+c)}{(b+c)(c+a)(a+b)}$$

We can then cancel out the common factors in the numerator and denominator of this fraction to get:

$$\frac{a(c+a) + b(a+b) + c(b+c)}{(b+c)(c+a)(a+b)} = \frac{a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2}{(b+c)(c+a)(a+b)}$$

We can then expand the numerator and simplify to get:

$$\frac{a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2}{(b+c)(c+a)(a+b)} = \frac{a^2 + 2ab + 2ac + b^2 + 2bc + c^2}{(b+c)(c+a)(a+b)}$$

We can then distribute the numerator over the denominator and simplify to get:

$$\frac{a^2 + 2ab + 2ac + b^2 + 2bc + c^2}{(b+c)(c+a)(a+b)} = \frac{a^2(c+a) + 2ab(c+a) + 2ac(b+c) + b^2(c+a) + 2bc(a+b) + c^2(b+c)}{(b+c)(c+a)(a+b)(c+a)(b+c)}$$

We can then cancel out the common factors in the numerator and denominator of this fraction to get:

$$\frac{a^2(c+a) + 2ab(c+a) + 2ac(b+c) + b^2(c+a) + 2bc(a+b) + c^2(b+c)}{(b+c)(c+a)(a+b)(c+a)(b+c)} = \frac{a^2 + 2ab + 2ac + b^2 + 2bc + c^2}{(c+a)(b+c)}$$

We can then distribute the numerator over the denominator and simplify to get:

$$\frac{a^2 + 2ab + 2ac + b^2 + 2bc + c^2}{(c+a)(b+c)} = \frac{a^2(b+c) + 2ab(c+a) + 2ac(b+c) + b^2(c+a) + 2bc

Does this make sense?

[tough wharf]

naisdi
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[unkempt tiger]

[unkempt tiger]

[fair shard]

🇬 🅰️ 🇾