At last an ineq USAMO '17 #6

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Atonu Roy Chowdhury
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At last an ineq USAMO '17 #6

Unread post by Atonu Roy Chowdhury » Sat Apr 22, 2017 9:32 am

Find the minimum possible value of $\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$ given that $a$, $b$, $c$, $d$ are nonnegative real numbers such that $a+b+c+d=4$.
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Thanic Nur Samin
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Re: At last an ineq USAMO '17 #6

Unread post by Thanic Nur Samin » Sat Apr 22, 2017 11:19 am

Here we use the well known tangent line trick. It is easy to see that the minimum is achieved at $(a,b,c,d)=(0,0,2,2)$ and its cyclic variants. Also, the most troublesome thing in this inequality is the quantities of the form $\dfrac{1}{x^3+4}$. So, we draw a tangent of $\dfrac{1}{x^3+4}$ at the point $x=2$.
USAMO6.png
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With a little calculation one can see that the equation of the tangent line is $-\dfrac{x}{12}+\dfrac{1}{4}$.

So, $\dfrac{1}{x^3+4}\ge-\dfrac{x}{12}+\dfrac{1}{4}$.

By using this, we get,

$\dfrac{a}{b^3+4}+\dfrac{b}{c^3+4}+\dfrac{c}{d^3+4}+\dfrac{d}{a^3+4}\ge 1-\dfrac{ab+bc+cd+da}{12}$.
Now, $\displaystyle ab+bc+cd+da=(a+c)(b+d)\le \left(\dfrac{a+b+c+d}{2}\right)^2=4$, So $\dfrac{a}{b^3+4}+\dfrac{b}{c^3+4}+\dfrac{c}{d^3+4}+\dfrac{d}{a^3+4}\ge \dfrac{2}{3}$
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Thanic Nur Samin
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Re: At last an ineq USAMO '17 #6

Unread post by Thanic Nur Samin » Sat Apr 22, 2017 11:24 am

For calculating the tangent, See that the derivative of the function at the point $2$ is $-\dfrac{1}{12}$, and plugging $2$ gives us $\dfrac{1}{12}$. So, the equation of the tangent is $-\dfrac{x}{12}+c$, and it goes through $(2,\dfrac{1}{12})$. This lets us determine the equation.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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Atonu Roy Chowdhury
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Re: At last an ineq USAMO '17 #6

Unread post by Atonu Roy Chowdhury » Sat Apr 22, 2017 2:00 pm

We'll show that $\sum_{cyc} \frac {a}{b^3 + 4} \ge \frac{2}{3} $
Subtracting $\frac{a}{4} + \frac{b}{4} + \frac{c}{4} + \frac{d}{4} = 1$ from both sides, we get
$\sum_{cyc} (\frac{a}{b^3 + 4} - \frac{a}{4}) \ge \frac{-1}{3}$
After some simplification, we get
$\sum_{cyc}\frac {3ab^3}{b^3 +4} \le 4$
By AM-GM, we get
$b^3 + 4 = \frac{b^3}{2} + \frac{b^3}{2} + 4 \ge 3b^2$ . So, $\sum_{cyc} \frac{3b^3}{b^3+4} \le \sum_{cyc} b $ with equality iff $b = 2$ or $b = 0$
It remains to prove that $\sum_{cyc} ab \le 4 $
$4 = (\frac{a+c+b+d}{2})^2 \ge (a+c)(b+d) = \sum_{cyc} ab $ and it can be obtained when $(a,b,c,d)=(2,2,0,0)$ and permutations.
This was freedom. Losing all hope was freedom.

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