here is my solution = let us denote $a,b,c$ as sides of triangle and $R$ as its circumradius than by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$ etc. so $AH.AD=4R^2cosAsinBsinC=cbcosA$ etc. than our inequality converts to proving that $ab+bc+ca\leq 2(cbcosA+abcosC+cacosB)$ now by cosine law ,we have $2cbcosA=b^2+c^2-a^2$ etc. substituting,our inequality becomes $ab+bc+ca\leq a^2+b^2+c^2$ which is obvious as it is equivalent to $\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\ge 0$ so we are done

Notice $ 2AH*AD=2AE*AC=2bc*cosA=c^2+b^2-a^2$ and so the problem translates to $a^2+b^2+c^2 \ge ab+bc+ca$, done.

You three had the exact same solution, and still posted it. A new approach, instead of using $cos$, we see that $2AH \cdot AD= AF \cdot AB + AE \cdot AC$, analogously $2BH \cdot BE= BF \cdot AB + BD \cdot BC$, and $2CH \cdot CF= CE \cdot AC + CD \cdot BC$, but $AF+BF=AB$, $AE+CE=AC$, and $CD+BD=BC$, so we need to prove $ab+bc+ca \le a^2+b^2+c^2$, which is well known.

not really contributing anything here, but might as well post my solution also, this seems too easy for an oly problem?

See here the proposed problem P2.

ABCDE wrote:

How does Power of a Point give you $AH\cdot AD=AF\cdot AB?$

programjames1 wrote: ABCDE wrote:

How does Power of a Point give you $AH\cdot AD=AF\cdot AB?$ Quadrilateral $HDBF$ is cyclic (from $\angle BFH = \angle HDB = 90$), and the sides $HD$ and $BF$ extended meet at $A$.

Note that it suffices to prove $2(AH\cdot AD+BH\cdot BE+CH\cdot CF)\geq ab+bc+ca.$ Claim 1: $AD\cdot AH=\frac{-a^2+b^2+c^2}{2}, BH\cdot BE=\frac{a^2-b^2+c^2}{2},$ and $CH\cdot CE=\frac{a^2+b^2-c^2}{2}$. Proof: Note that $AD=b\sin C$ and $CDHE$ is a cyclic quadrilateral. Then, $\angle AHE=\angle C$ so $AH=\frac{AE}{\sin C}=\frac{c\cos A}{\sin C}.$ Hence, $$AD\cdot AH=b\sin C\cdot \frac{c\cos A}{\sin C}=bc\cos A=bc(\frac{-a^2+b^2+c^2}{2bc})=\frac{-a^2+b^2+c^2}{2},$$as desired. The others follow similarly. $\Box$ By Claim 1, we simplify the equation to $2(\frac{-a^2+b^2+c^2}{2}+\frac{a^2-b^2+c^2}{2}+\frac{a^2+b^2-c^2}{2})=a^2+b^2+c^2\geq ab+bc+ca.$ This follows by the Trivial Inequality $$\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\geq 0.$$$\blacksquare$

Easy problem for Canado MO

is equivalent to $\sum{AH \cdot AD}$ = $\sum{bc \cdot \cos{A}}$ = $\frac{a^2 + b^2 + c^2}{2}$ Therefore the expression becomes $a^2 + b^2 + c^2 \ge ab + bc + ca$ which is trivial .

ABCDE wrote:

What is PoP in this context? Might be obvious but I simply can't make any connexions to any known results

That's pop applied to cylic quad $BFEC$

We have: $AB$ . $CA$ + $BC$ . $CA$ + $CA$ . $BC$ $\le$ $AB^2$ + $BC^2$ + $CA^2$ = $AF$ . $AB$ + $BF$ . $AB$ + $BD$ . $BC$ + $CD$ . $BC$ + $CE$ . $CA$ + $AE$ . $CA$ = 2 ($AH$ . $AD$ + $BH$ . $BE$ + $CH$ . $CF$) So: $\dfrac{AB . CA + BC . CA + CA . BC}{AH . AD + BH . BE + CH . CF}$ $\le$ 2 Equality holds when: $AB$ = $BC$ = $CA$ or $\triangle$ $ABC$ is equilateral

aditya21 wrote: here is my solution = let us denote $a,b,c$ as sides of triangle and $R$ as its circumradius than by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$ etc. so $AH.AD=4R^2cosAsinBsinC=cbcosA$ etc. than our inequality converts to proving that $ab+bc+ca\leq 2(cbcosA+abcosC+cacosB)$ now by cosine law ,we have $2cbcosA=b^2+c^2-a^2$ etc. substituting,our inequality becomes $ab+bc+ca\leq a^2+b^2+c^2$ which is obvious as it is equivalent to $\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\ge 0$ so we are done "by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$." I've been looking at this for like 2 days and I don't see how this is true... Could someone please provide some insight or hints.

By power of a point on cyclic quads $BDHF$ and $CDHE$, $AH\cdot AD=AF\cdot AB=AE\cdot AC$. Multiplying the denominator on the LHS to the RHS gives $$ \sum_{cyc} AB\cdot BC\leq \sum_{cyc} 2(AH\cdot AD)=\sum_{cyc} AF\cdot AB+AE\cdot AC=\sum_{cyc} AB^2 $$AM-GM completes the proof.

By PoP and AM-GM,\begin{align*}2(AH\cdot AC+BH\cdot BE+CH\cdot CF)&=AF\cdot AB+AE\cdot AC+BF\cdot BA+BD\cdot BC+CE\cdot CA+CD\cdot CB\\&=BC(CD+DB)+AC(DE+AE)+AB(AF+BF)\\&=BC^2+AC^2+AB^2\\&\ge AB\cdot AC+BC\cdot CA+CA\cdot CB\end{align*}$\square$

Let $a = \overline{BC}, b = \overline{CA}, c = \overline{AB}$. Without loss of generality, assume $a \leq b \leq c$. Observe that $AF = AC \cos \angle A$, and also $\triangle AFH \sim \triangle ADB$ so that \[\frac{AF}{AH} = \frac{AD}{AB} \implies \frac{AB}{AH} = \frac{AD}{AF} = \frac{AD}{AC \cos \angle A} \implies AH \cdot AD = \frac{b^2 + c^2 - a^2}{2}.\]Similarly, \[BH \cdot BE = \frac{c^2 + a^2 - b^2}{2}, \text{ and } \frac{a^2 + b^2 - c^2}{2}.\]Therefore our initial inequality is equivalent to \[\sum_{cyc}^{3} ab \leq \sum_{cyc}^{3} a^2, \]which is trivial by Chebyshev's rearrangement inequality. $\blacksquare$