Maths Olympiad Prep

Track / Stage 7 / 128 of 300 #1528 of 1964

Problem 1528

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.3 Prove it

Strengthening 1: If a,b,c∈R+a, b, c \in R_{+}, then
(∑ab+c)−32≥12⋅(a−b)2a2+b2+c2\left(\sum \frac{a}{b+c}\right)-\frac{3}{2} \geq \frac{1}{2} \cdot \frac{(a-b)^{2}}{a^{2}+b^{2}+c^{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.

Official solution

Prove: The original inequality is equivalent to
12[(c−a)2(a+b)(b+c)+(a−b)2(b+c)(c+a)+(b−c)2(c+a)(a+b)]≥12⋅(a−b)2a2+b2+c2⇔(c−a)2(a+b)(b+c)+(b−c)2(c+a)(a+b)≥[1a2+b2+c2−1(b+c)(c+a)](a−b)2\begin{array}{l} \frac{1}{2}\left[\frac{(c-a)^{2}}{(a+b)(b+c)}+\frac{(a-b)^{2}}{(b+c)(c+a)}+\frac{(b-c)^{2}}{(c+a)(a+b)}\right] \geq \frac{1}{2} \cdot \frac{(a-b)^{2}}{a^{2}+b^{2}+c^{2}} \\ \Leftrightarrow \frac{(c-a)^{2}}{(a+b)(b+c)}+\frac{(b-c)^{2}}{(c+a)(a+b)} \geq\left[\frac{1}{a^{2}+b^{2}+c^{2}}-\frac{1}{(b+c)(c+a)}\right](a-b)^{2} \end{array}

By the Cauchy-Schwarz inequality, we have:
(c−a)2(a+b)(b+c)+(b−c)2(c+a)(a+b)≥[(c−a)+(b−c)]2(a+b)(b+c)+(c+a)(a+b)=(a−b)2(a+b)(b+c)+(c+a)(a+b)\begin{array}{l} \frac{(c-a)^{2}}{(a+b)(b+c)}+\frac{(b-c)^{2}}{(c+a)(a+b)} \geq \frac{[(c-a)+(b-c)]^{2}}{(a+b)(b+c)+(c+a)(a+b)} \\ =\frac{(a-b)^{2}}{(a+b)(b+c)+(c+a)(a+b)} \end{array}

Therefore, it suffices to prove
1(a+b)(b+c)+(c+a)(a+b)≥1a2+b2+c2−1(b+c)(c+a)⇔c4+(a+b)c3−(a2+b2+3ab)c2+2(a3+b3−a2b−ab2)c+a4+b4+2(a3b+ab3)≥0\begin{array}{l} \frac{1}{(a+b)(b+c)+(c+a)(a+b)} \geq \frac{1}{a^{2}+b^{2}+c^{2}}-\frac{1}{(b+c)(c+a)} \\ \Leftrightarrow c^{4}+(a+b) c^{3}-\left(a^{2}+b^{2}+3 a b\right) c^{2}+2\left(a^{3}+b^{3}-a^{2} b-a b^{2}\right) c+a^{4}+b^{4}+2\left(a^{3} b+a b^{3}\right) \geq 0 \end{array}

Notice that
a3+b3−a2b−ab2=(a+b)(a2−ab+b2)−ab(a+b)=(a+b)(a−b)2≥0a^{3}+b^{3}-a^{2} b-a b^{2}=(a+b)\left(a^{2}-a b+b^{2}\right)-a b(a+b)=(a+b)(a-b)^{2} \geq 0

Therefore, it suffices to prove
c4+(a+b)c3−(a2+b2+3ab)c2+a4+b4+2(a3b+ab3)≥0c^{4}+(a+b) c^{3}-\left(a^{2}+b^{2}+3 a b\right) c^{2}+a^{4}+b^{4}+2\left(a^{3} b+a b^{3}\right) \geq 0

Since the above expression is homogeneous, we can assume c=1c=1, then the inequality can be transformed into:
a4+b4+2ab(a2+b2)−(a2+b2+3ab)+(a+b)+1≥0a^{4}+b^{4}+2 a b\left(a^{2}+b^{2}\right)-\left(a^{2}+b^{2}+3 a b\right)+(a+b)+1 \geq 0

Let a+b=u,ab=va+b=u, a b=v, then u2≥4vu^{2} \geq 4 v.
a2+b2=(a+b)2−2ab=u2−2va4+b4=(a2+b2)2−2a2b2=(u2−2v)2−2v2\begin{array}{l} a^{2}+b^{2}=(a+b)^{2}-2 a b=u^{2}-2 v \\ a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}=\left(u^{2}-2 v\right)^{2}-2 v^{2} \end{array}

Then the inequality is equivalent to:
(u2−2v)2−2v2+2v(u2−2v)−(u2−2v+3v)+u+1≥0⇔2v2+(2u2+1)v−u4+u2−u−1≤02v2+(2u2+1)v−u4+u2−u−1≤2(u24)2+(2u2+1)⋅u24−u4+u2−u−1=−3u4−10u2+u−18=−(u+2)(3u3−6u2+2u+4)8=−(u+2)[(12u3+12u3+4−3u2)+2(u3+u−2u2)+u2]8=−(u+2)[(12u3+12u3+4−3u2)+2(u3+u−2u2)+u2]8=−(u+2)[(2123u+43)(123u−43)2+2u(u−1)2+u2]8≤0\begin{array}{l} \left(u^{2}-2 v\right)^{2}-2 v^{2}+2 v\left(u^{2}-2 v\right)-\left(u^{2}-2 v+3 v\right)+u+1 \geq 0 \\ \Leftrightarrow 2 v^{2}+\left(2 u^{2}+1\right) v-u^{4}+u^{2}-u-1 \leq 0 \\ 2 v^{2}+\left(2 u^{2}+1\right) v-u^{4}+u^{2}-u-1 \\ \leq 2\left(\frac{u^{2}}{4}\right)^{2}+\left(2 u^{2}+1\right) \cdot \frac{u^{2}}{4}-u^{4}+u^{2}-u-1 \\ =-\frac{3 u^{4}-10 u^{2}+u-1}{8} \\ =-\frac{(u+2)\left(3 u^{3}-6 u^{2}+2 u+4\right)}{8} \\ =-\frac{(u+2)\left[\left(\frac{1}{2} u^{3}+\frac{1}{2} u^{3}+4-3 u^{2}\right)+2\left(u^{3}+u-2 u^{2}\right)+u^{2}\right]}{8} \\ =-\frac{(u+2)\left[\left(\frac{1}{2} u^{3}+\frac{1}{2} u^{3}+4-3 u^{2}\right)+2\left(u^{3}+u-2 u^{2}\right)+u^{2}\right]}{8} \\ =-\frac{(u+2)\left[\left(2 \sqrt[3]{\frac{1}{2}} u+\sqrt[3]{4}\right)\left(\sqrt[3]{\frac{1}{2}} u-\sqrt[3]{4}\right)^{2}+2 u(u-1)^{2}+u^{2}\right]}{8} \leq 0 \end{array}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.