Maths Olympiad Prep

Track / Stage 8 / 20 of 180 #1720 of 1964

Problem 1720

IMO Shortlist mid-range; USAMO P2/P5
Algebra Difficulty 8.1 Prove it

Let 0⩽a,b,c⩽10 \leqslant a, b, c \leqslant 1, then
a1+b+c+b1+c+a+c1+a+b+(1−a)(1−b)(1−c)⩾78\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}+(1-a)(1-b)(1-c) \geqslant \frac{7}{8}

Equality in (65) holds if and only if a=b=c=12a=b=c=\frac{1}{2}.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

 The left side of equation (65) - the right side =∑ab+c+1+1−∑a+∑bc−abc−78=(12∑bc−34∑a+∑ab+c+1)+(18−14∑a+12∑bc−abc)=[14∑a(b+c+1)−∑a+∑ab+c+1]+(12−a)(12−b)(12−c)=∑a(b+c+1−2)24(b+c+1)+(12−a)(12−b)(12−c)=∑a(b+c−1)24(b+c+1)+(12−a)(12−b)(12−c)\begin{array}{l} \text { The left side of equation (65) - the right side }=\sum \frac{a}{b+c+1}+1-\sum a+\sum b c-a b c-\frac{7}{8}= \\ \left(\frac{1}{2} \sum b c-\frac{3}{4} \sum a+\sum \frac{a}{b+c+1}\right)+\left(\frac{1}{8}-\frac{1}{4} \sum a+\frac{1}{2} \sum b c-a b c\right)= \\ {\left[\frac{1}{4} \sum a(b+c+1)-\sum a+\sum \frac{a}{b+c+1}\right]+\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right)=} \\ \sum \frac{a(b+c+1-2)^{2}}{4(b+c+1)}+\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right)= \\ \sum \frac{a(b+c-1)^{2}}{4(b+c+1)}+\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right) \end{array}

From this, to prove that equation (65) holds, we only need to prove:
∑a(b+c−1)24(b+c+1)+(12−a)(12−b)(12−c)⩾0\sum \frac{a(b+c-1)^{2}}{4(b+c+1)}+\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right) \geqslant 0

We will prove this in two cases.
i) If (12−a)(12−b)(12−c)⩾0\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right) \geqslant 0, then inequality (1) is obviously true, with equality holding if and only if a=b=c=12a=b=c=\frac{1}{2}.
ii) If (12−a)(12−b)(12−c)<0\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-c\right) < 0, by symmetry, we can assume 12−a⩽12−b⩽12−c\frac{1}{2}-a \leqslant \frac{1}{2}-b \leqslant \frac{1}{2}-c. We will further divide this into two sub-cases.
i) If a−12⩾b−12⩾c−12>0a-\frac{1}{2} \geqslant b-\frac{1}{2} \geqslant c-\frac{1}{2}>0, then
a⩾2(a−12)⩾0,(b+c−1)2⩾4(b−12)(c−12)>0a \geqslant 2\left(a-\frac{1}{2}\right) \geqslant 0, \quad (b+c-1)^{2} \geqslant 4\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right)>0

So
a(b+c−1)24⩾2(a−12)(b−12)(c−12)\frac{a(b+c-1)^{2}}{4} \geqslant 2\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right)

Also
b+c+1⩽3b+c+1 \leqslant 3

So
a(b+c−1)24(b+c+1)⩾23(a−12)(b−12)(c−12)\frac{a(b+c-1)^{2}}{4(b+c+1)} \geqslant \frac{2}{3}\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right)

Noting the conditions for equality in the above inequalities, we know that the equality cannot hold, thus
a(b+c−1)24(b+c+1)>23(a−12)(b−12)(c−12)\frac{a(b+c-1)^{2}}{4(b+c+1)}>\frac{2}{3}\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right)

Similarly, we can obtain the other two inequalities. Adding these three inequalities on both sides, we get
∑a(b+c−1)24(b+c+1)>2(a−12)(b−12)(c−12)⩾(a−12)(b−12)(c−12)\begin{array}{l} \sum \frac{a(b+c-1)^{2}}{4(b+c+1)}>2\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right) \geqslant \\ \left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right) \end{array}

Thus, we obtain inequality (1).
ii) If a−12>0>b−12⩾c−12a-\frac{1}{2}>0>b-\frac{1}{2} \geqslant c-\frac{1}{2}, then
a⩾2(a−12),(b+c−1)2⩾4(b−12)(c−12)a \geqslant 2\left(a-\frac{1}{2}\right), \quad (b+c-1)^{2} \geqslant 4\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right)

Also
b+c+1⩽2b+c+1 \leqslant 2

So
a(b+c−1)24(b+c+1)>(a−12)(b−12)(c−12)\frac{a(b+c-1)^{2}}{4(b+c+1)}>\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right)

Thus
∑a(b+c−1)24(b+c+1)>(a−12)(b−12)(c−12)\sum \frac{a(b+c-1)^{2}}{4(b+c+1)}>\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right)

We also obtain inequality (1).
In summary, equation (65) is proven, with equality holding if and only if a=b=c=12a=b=c=\frac{1}{2}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.