v_Enhance wrote: A triangle has side lengths $a$, $b$, $c$, and the altitudes have lengths $h_a$, $h_b$, $h_c$. Prove that \[ \left( \frac{a}{h_a} \right)^2 + \left( \frac{b}{h_b} \right)^2 + \left( \frac{c}{h_c} \right)^2 \ge 4. \] The inequality is equivalent to \[a^4+b^4+c^4\geq 16S^2.\] here $S$ be area of the triangle. (This inequality can be strengthened.) here here

Let $K$ denote the area of triangle $ABC$. We clearly have $h_A = \frac{2K}{a}$, etc. so our inequality becomes \begin{align*}a^4 + b^4 + c^4 \ge 16K^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \\ &= -a^4 -b^4 - c^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2\end{align*} We thus have the inequality $a^4 + b^4 + c^4 \ge a^2b^2 + b^2c^2 + c^2a^2$ which can be proven in a variety of ways, like: \[\sum_{cyc} (a^2-b^2)^2 \ge 0\]

Using $h_a=\frac{2S}{a}$, $h_b=\frac{2S}{b}$ and $h_c=\frac{2S}{c}$ inequality is equvivalent to $a^4+b^4+c^4 \geq 16S^2 $. With $CBS$ and this inequality we get $a^4+b^4+c^4 \geq \frac{(a^2+b^2+c^2)^2}{3} \geq 16S^2 $ q.e.d. Equality is achieved if and only if $a=b=c$

$a^4+b^4+c^4 \geq \frac{(a^2+b^2+c^2)^2}{3} \geq 16S^2 .$ Good.

Letting $ h_a = \frac{2K}{a} $ and $ h_b = \frac{2K}{b} $ and $ h_c = \frac{2K}{c} $ where $ K $ is the area of triangle $ \triangle{ABC} $ we have by Cauchy-Schwartz that $ \sum_{cyc}\left(\frac{a}{h_a}\right)^2 = \sum_{cyc}\left(\frac{a^2}{2K}\right)^2 \ge \frac{1}{3}\left(\sum_{cyc}\frac{a^2}{2K}\right)^2 $ so it suffices to show that $ a^2 + b^2 + c^2 \ge 4\sqrt{3}K $ which is a weaker case of the Hadwiger-Finsler inequality.

Since $ah_a=bh_b=ch_c=2K$, where $K$ is the area of the given triangle, it suffices to prove that \[a^4+b^4+c^4\ge 16K^2.\] However, by Heron's formula, we have \[16K^2 = \left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right),\] so the original inequality is actually equivalent to \[3\left(a^4+b^4+c^4\right)\ge \left(a^2+b^2+c^2\right)^2\] which is true by Cauchy-Schwarz$.\;\blacksquare$

Ravi substitution does the trick. Just substitute: a=y+z b=z+x c=x+y And see the magic ! A.W.D

sqing wrote: $a^4+b^4+c^4 \geq \frac{(a^2+b^2+c^2)^2}{3} \geq 16S^2 .$ Good. $a^4+b^4+c^4 \geq 16S^2+(b^2+c^2-2a^2)^2 .$ http://www.artofproblemsolving.com/community/c6h487317p2730515

My solution: Lemma: $(x+y)^4+(y+z)^4+(z+x)^4\geq 16xyz(x+y+z)$ Proof: $(x+y)^4+(y+z)^4+(z+x)^4\geq \frac{(2x+2y+2z)^4}{27}= \frac{16(x+y+z)(x+y+z)^3}{27}\geq 16xyz(x+y+z)$ In the problem: Replace: $S^2=xyz(x+y+z)$ and $a=x+y,b=y+z,c=z+x$ The inequality is equivalent to $a^4+b^4+c^4 \geq 16S^2$ if only if $(x+y)^4+(y+z)^4+(z+x)^4\geq 16xyz(x+y+z)$ which it is true by lemma. $\Longrightarrow$ $\frac{a^2}{h_a^2}+ \frac{b^2}{h_b^2}+\frac{c^2}{h_c^2}\geq 4$...

Hm, that was pretty easy for a TST question? Let $A$ be the area of the triangle and $a=x+y,b=y+z$ and $c=z+x$ by Ravi. Using Heron, we'd get $F=\sqrt{xyz(x+y+z)}$. Now we also have $h_a=\tfrac{2F}{a}$ and similar expressions for $b,c$. Thus, the inequality is equivalent to \[ (x+y)^4+(y+z)^4+(z+x)^4 \geq 16 xyz(x+y+z). \]Luckily, that is easy to prove via AM-GM. We have \[ \sum_{\text{cyc}} (x+y)^4 \geq \sum_{\text{cyc}} 16x^2y^2 = 16 \sum_{\text{cyc}} \frac{x^2y^2+z^2x^2}{2} \geq 16 \sum_{\text{cyc}} x^2yz \]Done. $\hfill \blacksquare$

Using sine law we get that it's equivalent to $$(\frac{2aR}{bc})^{2} +(\frac{2bR}{ac})^{2} + (\frac{2cR}{ab})^{2} \ge 4$$which is equivalent to $$R^{2}(a^{4} + b^{4} + c^{4}) \ge a^{2}b^{2}c^{2}$$where $R$ is the radius of the circumcircle. Using the well-known inequality that $R^{2} \ge \frac{1}{9} (a^{2} + b^{2} + c^{2})$ and $AM-GM$ inequality we have $$ R^{2}(a^{4} + b^{4} + c^{4}) \ge \frac{1}{9} \cdot(a^{2} + b^{2} + c^{2})(a^{4} + b^{4} + c^{4}) \ge \frac{1}{9} 3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}\cdot 3a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}} = a^{2}b^{2}c^{2}$$and we are done.

Quite easy problem for Taiwan.

The inequality reduces to proving that $(\frac{sinA}{sinBsinC})^2+(\frac{sinB}{sinAsinC})^2+(\frac{sinC}{sinAsinB})^2\geq 4$....(1) We have used the fact that $a=2RsinA$ and $h_a=2RsinBsinC$ and symmetric for others. Now by using $a^2+b^2+c^2\geq ab+bc+ca$, we get that the LHS of (1) is $\geq (\frac{1}{sinC}^2+(\frac{1}{sinB})^2+(\frac{1}{sinC})^2$.......(2) Now we apply the same inequality again and get that (2) is$\geq \frac{1}{sinAsinB}+\frac{1}{sinBsinC}+\frac{1}{sinCsinA}$.........(3) Now we prove that (3) is $\geq 4$ .We get that (3) is equal to $\frac{sinA+sinB+sinC}{sinAsinBsinC}=\frac{4cos(\frac{A}{2})cos(\frac{B}{2})cos(\frac{C}{2})}{sinAsinBsinC}$ By using $sin(2A)=2sinAcosA$, the expression simplifies to $\frac{1}{2sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})}$ By AM-GM and Jensen inequality on the function $f(x)=sinx$ we get that the denominator is less than or equal to $\frac{1}{4}$ so its reciprocal is greater than this quantity so we get our desired result.

Took me <5 mins Clearly $h_a=\dfrac{2\Delta}{a}$ so we have to show \begin{align*}a^4+b^4+c^4 &\ge 16\Delta^2 \\ &= (a+b+c)(a+b-c)(c+(a-b))(c-(a-b)) \\ &= ((a+b)^2-c^2)(c^2-(a-b)^2) \\ &= c^2((a+b)^2+(a-b)^2)-c^4-(a^2-b^2)^2 \\ &= 2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4 \end{align*}So it suffices to show $a^4+b^4+c^4 \ge a^2b^2+b^2c^2+c^2a^2$ which is clear.$\blacksquare$

Let $K$ be the area of the triangle. $ah_a = 2K \implies \frac{a}{h_a} = \frac{a^2}{2K}$ so the given is equivalent to $\sum a^4 \ge 16K^2$. Note \[ 16K^2 = 16s(s-a)(s-b)(s-c) = ((b+c)^2-a^2)(a^2-(b-c)^2) = 2\sum a^2b^2 - \sum a^4 \]So it suffices to prove $\sum a^4 \ge \sum a^2b^2$ which is Muirhead.

Simply transfer the problem to the $\tan$-inequality. Indeed, we have $\tan{A}+\tan{B}+\tan{C}=\tan{A}\cdot \tan{B}\cdot \tan{C}$, where $A+B+C=\pi$ and actually inequality is equivalent to $$\sum_{cyc}\left(\frac{1}{\tan{B}}+\frac{1}{\tan{C}}\right)^2\geq 4.$$For clarity, let $x=\tan{A},y=\tan{B},z=\tan{C}$. Thus, we need to show that $$\sum_{cyc}\left(\frac{1}{y}+\frac{1}{z}\right)^2\geq 4$$under constraint $xyz=x+y+z$. Indeed, \begin{align*} \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}&\geq 2\Longleftrightarrow\\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}&\geq 1\Longleftrightarrow\\ x^2y^2+y^2z^2+z^2y^2&\geq x^2y^2z^2=xyz(x+y+z), \end{align*}which is true by Muirhead. $\blacksquare$