An Universal Inequality for Cevians

Let in $\Delta ABC,$ $p=\min\{a,b,c\},$ $q=\max\{a,b,c\};$ $M\in BC,$ $N\in AC,$ $P\in AB.$ Prove that

$\displaystyle 3+\frac{9pq}{AM^2+BN^2+CP^2}\le (p+q)\left(\frac{1}{AM}+\frac{1}{BN}+\frac{1}{CP}\right).$


For $l_a,\,l_b,\,l_c$ the angle bisectors in $\Delta ABC,$

$\displaystyle 3+\frac{9pq}{l_a^2+l_b^2+l_c^2}\le (p+q)\left(\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c}\right).$

For $m_a,\,m_b,\,m_c$ the medians in $\Delta ABC,$

$\displaystyle 3+\frac{9pq}{m_a^2+m_b^2+m_c^2}\le (p+q)\left(\frac{1}{m_a}+\frac{1}{m_b}+\frac{1}{m_c}\right).$


Naturally, $AM,BN,CP\in[ p,q]$ so that $(p-AM)(q-AM)\le 0,$ implying

$pq-(p+q)AM+AM^2\le 0,$

which is the same as


Similar inequalities hold for $BN$ and $CP.$ Taking the sum of all three gives

$\displaystyle 3+pq\sum_{cyc}\frac{1}{AM^2}\le (p+q)\sum_{cyc}\frac{1}{AM}.$

By the Harmonic Mean - Arithmetic Mean Inequality,

$\displaystyle \frac{9}{\displaystyle\sum_{cyc}AM^2}\le\sum_{cyc}\frac{1}{AM^2}$

such that

$\displaystyle 3+\frac{9pq}{\displaystyle\sum_{cyc}AM^2}\le 3+pq\sum_{cyc}\frac{1}{AM^2}.$

By the transitivity of the relation of inequality,

$\displaystyle 3+\frac{9pq}{AM^2+BN^2+CP^2}\le (p+q)\left(\frac{1}{AM}+\frac{1}{BN}+\frac{1}{CP}\right).$


The above is a fallacy invented by Dan Sitaru and Leo Giugiuc and posted at the CutTheKnotMath facebook page. As a hint of what is wrong with the proof, note that it would have been nice to mention the applicability of the inequality to the altitudes $h_a,\,h_b,\,h_c.$ The omission was deliberate!

Another way to see that there is something wrong with the above is to consider the most symmetric configuration of an equilateral triangle in which $M,\,N,\,P$ are the midpoints of the corresponding sides. The inequality then appears to assert that $7\le 4\sqrt{3}.$

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