Tessellate S.T.E.M.S. (2019) - Mathematics - School - Set 1 - Subjective Problem 2

Given three positive real numbers \(a,b,c\) satisfying \(abc=1\). Prove or disprove (i.e. find a counterexample) to the following inequality \[\sum_{cyc} \frac{1}{a(a+1)+ab(ab+1)} \geq \frac{3}{4}\].

This problem is a part of Tessellate S.T.E.M.S. (2019)

#Algebra

Note by Tessellate S.T.E.M.S. Mathematics
2 years, 8 months ago

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Solution:

We claim that the given inequality is true for all positive real numbers a,b,c a, b, c satisfying abc=1 abc = 1. For this we consider the positive real numbers x:=ab=1c,y:=1,z:=a x:=ab=\frac{1}{c}, y:=1, z:=a which satisfy a=zy,b:=xz,c:=yxa = \frac{z}{y}, b:= \frac{x}{z}, c:= \frac{y}{x} so that ∑cyc1a(a+1)+ab(ab+1)=∑cycy2z(z+y)+x(x+y)\sum_{cyc} \frac{1}{a(a+1)+ab(ab+1)} = \sum_{cyc} \frac{y^2}{z(z+y)+x(x+y)} .

Now we observe that by the Cauchy-Schwarz Inequality, (∑cycy2z(z+y)+x(x+y))(∑cycy2(z2+zy+x2+xy))≥(∑cycy2)2(\sum_{cyc} \frac{y^2}{z(z+y)+x(x+y)}) (\sum_{cyc} y^2(z^2+zy+x^2+xy)) \geq (\sum_{cyc} y^2)^2

which yields (∑cycy2z(z+y)+x(x+y))≥(∑cycy2)2∑cycy2(z2+zy+x2+xy)(\sum_{cyc} \frac{y^2}{z(z+y)+x(x+y)}) \geq \frac{(\sum_{cyc} y^2)^2}{\sum_{cyc} y^2(z^2+zy+x^2+xy)}

Finally, we note that since 4(∑cycy2)2−3∑cycy2(z2+zy+x2+xy)=4∑cycy4+2∑cycx2y2−3∑cycxy(x2+y2)=3∑cyc(x−y)2(x2+y2)+∑cyc(x2−y2)2≥04(\sum_{cyc} y^2)^2 - 3\sum_{cyc} y^2(z^2+zy+x^2+xy) = 4\sum_{cyc} y^4 + 2\sum_{cyc} x^2y^2 - 3\sum_{cyc} xy(x^2+y^2) = 3\sum_{cyc} (x-y)^2(x^2+y^2) + \sum_{cyc}(x^2-y^2)^2 \geq 0

we obtain (∑cycy2)2∑cycy2(z2+zy+x2+xy)≥34\frac{(\sum_{cyc} y^2)^2}{\sum_{cyc} y^2(z^2+zy+x^2+xy)} \geq \frac{3}{4}

which yields the desired result.

Tessellate S.T.E.M.S. Mathematics - 2 years, 7 months ago

Direct Titu's Lemma.

Saikat Sengupta - 2 years, 6 months ago
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