$a, b, c$ are distinct therefore equality cannot be achieved if the expression is equal to $a+b+c$

svatejas wrote: 2. Let \(a,b,c\) be three distinct positive real numbers such that \(abc=1\). Prove that \(\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)} \ge 3\). The following inequality is also true. Let \(a,b,c\) be three distinct positive real numbers such that \(abc=1\). Prove that \[\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)} \le \frac{(ab+bc+ca)(a+b+c)}{3}\]

When is equality achieved?

NO equality $\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}=a+b+c>3$

oldbeginner wrote: svatejas wrote: 2. Let \(a,b,c\) be three distinct positive real numbers such that \(abc=1\). Prove that \(\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)} \ge 3\). The following inequality is also true. Let \(a,b,c\) be three distinct positive real numbers such that \(abc=1\). Prove that \[\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)} \le \frac{(ab+bc+ca)(a+b+c)}{3}\] note that $\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}=a+b+c$

Is the following solution correct? WLOG $a>b>c$. Multiplying both sides by $(a-b)(b-c)(c-a)$, it suffices to prove $\sum a^3(b-c) \ge 3(a-b)(b-c)(c-a)$. Expanding both sides and simplifying , this reduces to prove $\sum (a^3b+3a^2b) \ge \sum( ab^3+3ab^2)$. Using $abc=1$ this simplifies to show $$\sum \frac{a^2+3a}{c} \ge \sum \frac{a^2+3a}{b}.$$ But this is just rearrangement on $a^3+3a>b^3+3b>c^3+3c$ and $1/a < 1/b < 1/c$.

I assumed that a>b>c or a>c>b since the equality is cyclic and not symmetric.

perhaps the most elegant solution is the following We interpolate at $a^3,b^3,c^3$ ( because by wishful thinking and AM GM on $a+b+c$) define the polynomial $P(t)= \sum \frac {a^3(t-b)(t-c)}{(a-b)(a-c)}$ Note that $P(t)=t^3$ for $t=a,b,c$ so $P(t) -t^3=k(t-a)(t-b)(t-c)$ so $k=-1$ as $P(t)$ is a two degree polynomial compare the coefficient of $t^2$ to find that the sum in the question is equal to only $a+b+c$(Mathemagic) so by AM GM on $a+b+c$ result follows.

luofangxiang wrote: How would you get this?

DaniyalQazi2 wrote: How would you get this? Getting a common denominator, the LHS of the original inequality becomes $\frac{ -a^3b + a^3c -b^3c + b^3a-c^3a + c^3b}{(a-b)(b-c)(c-a) }$. Now, notice that if $a=b$, the numerator becomes $-a^4 + a^3c - a^3c + a^4 - c^3a + c^3 a = 0$, so $(a-b)$ is a root of the numerator, and similarly so are $(b-c)$ and $(c-a)$. Now, let's try to factor out an $a-b$ from the numerator. We get $ab(b^2-a^2) - c^3 ( a-b) + c(a^3-b^3)$, using factor by grouping, which is equal to $(a-b)(-ab(a+b) - c^3 + c(a^2+ab+b^2))$. Thus, our fraction becomes $\frac{-a^2b-ab^2 -c^3 + ca^2 + abc + cb^2}{(b-c)(c-a)}$. Then, we try to factor $(b-c)$ out of the numerator. This gives $c(b^2 - c^2) - ab(b-c) -a^2 ( b-c) = (b-c)(bc+c^2 -ab - a^2)$. Our fraction is now $\frac{bc+c^2 - ab - a^2}{c-a} = \frac{(c-a)(a+b+c)}{c-a} = a+b+c$, as luofangxiang claimed.

In my method above you can find an elegant process(shameless showoff).

As a,b, c>0, a-b<=a => 1\(a-b)>=1\a Similarly, 1\(a-c)>=1\a 1\(b-c)>=1\b 1\(b-a)>=1\b 1\(c-a)>=1\c 1\(c-b)>=1\c Therefore, a^3\(a-b)(a-c) >= a^3\a^2 = a Similar case for the other two So, enough to prove a+b+c>=3 which can be easily proved by AM-GM

#atmchallege's solution was perhaps the easiest and fastest

svatejas wrote: Let \(a,b,c\) be three distinct positive real numbers such that \(abc=1\). Prove that $$\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)} \ge 3$$ https://olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/11/3sol.pdf

Attachments:

vjdjmathaddict wrote: perhaps the most elegant solution is the following We interpolate at $a^3,b^3,c^3$ ( because by wishful thinking and AM GM on $a+b+c$) define the polynomial $P(t)= \sum \frac {a^3(t-b)(t-c)}{(a-b)(a-c)}$ Note that $P(t)=t^3$ for $t=a,b,c$ so $P(t) -t^3=k(t-a)(t-b)(t-c)$ so $k=-1$ as $P(t)$ is a two degree polynomial compare the coefficient of $t^2$ to find that the sum in the question is equal to only $a+b+c$(Mathemagic) so by AM GM on $a+b+c$ result follows. What are you comparing the coefficient of $t^2$ with?

svatejas wrote: Let \(a,b,c\) be three distinct positive real numbers such that \(abc=1\). Prove that $$\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)} \ge 3$$ Let \(a,b,c\) be distinct binary real number such that \(a+b+c=3\). Prove that $$\frac{a^4} {(a-b)(a-c)} +\frac{b^4} {(b-c)(b-a)}+\frac{c^4} {(c-a)(c-b)}>6$$

Bash Bash and Bash !!! Let a³/(a - b)(a - c) + b³/(b - c)(b - a) + c³/(c - a)(c - b) = t where a,b and c are three distinct positive real numbers such that abc = 1. t = -[a³(b - c) + b³(c - a) + c³(a - b)]/(a - b)(b - c)(c - a) Now, a³(b - c) + b³(c - a) + c³(a - b) = a³b - ab³ + b³c - ca³ + c³a - bc³ = ab(a² - b²) - c(a³ - b³) + c³(a - b) = (a - b)[ab(a + b) - c(a² + ab + b²) + c³] = (a - b)[a²b + ab² - ca² - abc - b²c + c³] = (a - b)[a²(b - c) + ab(b - c) - c(b² - c²)] = (a - b)(b - c)[a² + ab - c(b + c)] = (a - b)(b - c)[a² + ab - bc - c²)] = (a - b)(b - c)[-b(c - a) - (c² - a²)] = -(a - b)(b - c)(c - a)(a + b + c) So, t = a + b + c Then, by AM-GM Inequality we've t ≥ 3(abc)^⅓ ≥ 3 Therefore, t ≥ 3 and we're done.