Maths Olympiad Prep

Library / /110 of 520

Algebra Difficulty 6.1 National olympiad Prove it

Example 1.2.4. Let a,b,ca, b, c be positive real numbers with sum 3. Prove that
a2a+2b2+b2b+2c2+c2c+2a2≥1 \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1

Solution

Solution. We use the following estimation according to AM-GM
a2a+2b2=a−2ab2a+2b2≥a−2ab23ab43=a−2(ab)2/33\frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{2 / 3}}{3}
which implies that
∑cyc a2a+2b2≥∑cyc a−23∑cyc (ab)23\sum_{\text {cyc }} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{\text {cyc }} a-\frac{2}{3} \sum_{\text {cyc }}(a b)^{\frac{2}{3}}

It suffices to prove that
(ab)2/3+(bc)2/3+(ca)2/3≤3.(a b)^{2 / 3}+(b c)^{2 / 3}+(c a)^{2 / 3} \leq 3 .

By AM-GM, we have the desired result since
3∑cyca≥2∑cyca+∑cycab=∑cyc(a+b+ab)≥3∑cyc(ab)23.3 \sum_{c y c} a \geq 2 \sum_{c y c} a+\sum_{c y c} a b=\sum_{c y c}(a+b+a b) \geq 3 \sum_{c y c}(a b)^{\frac{2}{3}} .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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