Ulikhetmaraton
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UTAG anta at $a\leq c\leq b$. Vi viser at $\ell$ lik vinkelhalveringslinjen tilfredsstiller ulikheten i oppgaven. La $D=\ell \cap BC$. Snittet av $ABC$ og $A'B'C'$ er lik firkanten $ABDB'$. Videre er arealene $[ABD]=[ADB']=\frac{c}{b+c}[ABC]$ av vinkelhalveringslinjesetningen. Det holder dermed å vise at $\frac{c}{b+c}>\frac{1}{3}$, men det følger av at $b<a+c\leq 2c$.
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nilpotent1
- Pytagoras
- Posts: 6
- Joined: 30/09-2025 14:52
Satisfying inequality 
Proof: Notice,
\[
\sum_{\text{cyc}} \left( \frac{1}{a + b + \sqrt{2a + 2c}} \right)^3 = \sum_{\text{cyc}} \left( \frac{1}{a + b + 2 \cdot \sqrt{\frac{a + c}{2}} } \right)^3
\]
however by AM-GM,
\[
\sum_{\text{cyc}} \left( \frac{1}{a + b + 2 \cdot \sqrt{\frac{a + c}{2}} } \right)^3 \leq \sum_{\text{cyc}} \left( \frac{1}{3 \sqrt[3]{\frac{(a + b)(a + c)}{2}}} \right)^3 = \frac{1}{3^3} \left( \sum_{\text{cyc}} \frac{2}{(a + b)(a + c)} \right) \overset{?}{\leq} \frac{8}{9}
\]
thus we want to show that,
\[
\sum_{\text{cyc}} \frac{1}{(a + b)(a + c)} \leq 12
\]
multiplying by $(a + b)(b + c)(c + a)$ we obtain,
\[
2(a + b + c) \leq 12 (a + b)(b + c)(c + a) = 12(2abc + a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b) = 12 \biggl[ (ab + bc + ca)(a + b + c) - abc \biggr]
\]
from the problem statement we know that $16abc(a + b + c) \geq ab + bc + ca$ let us multiply both sides of the desired result by it to homogenize,
\[
6 \biggl[ (ab + bc + ca)(a + b + c) - abc \biggr] (ab + bc + ca) \overset{?}{\geq} 16abc(a + b + c)^2
\]
\[
3(ab + bc + ca)^2(a + b + c) \overset{?}{\geq} 8abc(a + b + c)^2 + 3abc(ab + bc + ca)
\]
Notice that $abc(a + b + c) \leq \frac{1}{3} (ab + bc + ca)^2$ (trivial by Rearrangment inequality for example), thus,
\[
8abc(a + b + c)^2 + 3abc(ab + bc + ca) \leq \frac{8}{3} (ab + bc + ca)^2 (a + b + c) + 3abc(ab + bc + ca) \overset{?}{\leq} 3(ab + bc + ca)^2 (a + b + c)
\]
thus all we want to show is that
\[
9 abc(ab + bc + ca) \leq (ab + bc + ca)^2 (a + b + c)
\]
\[
9 abc \leq (ab + bc + ca) (a + b + c)
\]
and the last inequality is true due to AM-GM. $\blacksquare$
Proof: Notice,
\[
\sum_{\text{cyc}} \left( \frac{1}{a + b + \sqrt{2a + 2c}} \right)^3 = \sum_{\text{cyc}} \left( \frac{1}{a + b + 2 \cdot \sqrt{\frac{a + c}{2}} } \right)^3
\]
however by AM-GM,
\[
\sum_{\text{cyc}} \left( \frac{1}{a + b + 2 \cdot \sqrt{\frac{a + c}{2}} } \right)^3 \leq \sum_{\text{cyc}} \left( \frac{1}{3 \sqrt[3]{\frac{(a + b)(a + c)}{2}}} \right)^3 = \frac{1}{3^3} \left( \sum_{\text{cyc}} \frac{2}{(a + b)(a + c)} \right) \overset{?}{\leq} \frac{8}{9}
\]
thus we want to show that,
\[
\sum_{\text{cyc}} \frac{1}{(a + b)(a + c)} \leq 12
\]
multiplying by $(a + b)(b + c)(c + a)$ we obtain,
\[
2(a + b + c) \leq 12 (a + b)(b + c)(c + a) = 12(2abc + a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b) = 12 \biggl[ (ab + bc + ca)(a + b + c) - abc \biggr]
\]
from the problem statement we know that $16abc(a + b + c) \geq ab + bc + ca$ let us multiply both sides of the desired result by it to homogenize,
\[
6 \biggl[ (ab + bc + ca)(a + b + c) - abc \biggr] (ab + bc + ca) \overset{?}{\geq} 16abc(a + b + c)^2
\]
\[
3(ab + bc + ca)^2(a + b + c) \overset{?}{\geq} 8abc(a + b + c)^2 + 3abc(ab + bc + ca)
\]
Notice that $abc(a + b + c) \leq \frac{1}{3} (ab + bc + ca)^2$ (trivial by Rearrangment inequality for example), thus,
\[
8abc(a + b + c)^2 + 3abc(ab + bc + ca) \leq \frac{8}{3} (ab + bc + ca)^2 (a + b + c) + 3abc(ab + bc + ca) \overset{?}{\leq} 3(ab + bc + ca)^2 (a + b + c)
\]
thus all we want to show is that
\[
9 abc(ab + bc + ca) \leq (ab + bc + ca)^2 (a + b + c)
\]
\[
9 abc \leq (ab + bc + ca) (a + b + c)
\]
and the last inequality is true due to AM-GM. $\blacksquare$
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nilpotent1
- Pytagoras
- Posts: 6
- Joined: 30/09-2025 14:52
Ny Oppgave: Vis at for alle $a, b, c \in \mathbb{R}^{+}$ har vi,
\[
\sum_{\text{cyc}} \frac{\sqrt{b + c}}{a} \geq \frac{4(a + b + c)}{\sqrt{(a + b)(b + c)(c + a)}}
\]
\[
\sum_{\text{cyc}} \frac{\sqrt{b + c}}{a} \geq \frac{4(a + b + c)}{\sqrt{(a + b)(b + c)(c + a)}}
\]
Vi skriver om ulikheten i oppgaven til
\[
\sum_{cyc} \frac{b+c}{a}\sqrt{(a+b)(a+c)} \geq 4(a+b+c)
\]
AV CS på $\sqrt{(a+b)(a+c)}$ har vi $\frac{b+c}{a}\sqrt{(a+b)(a+c)}\geq \frac{b+c}{a}\sqrt{bc} + b+c$.
Det holder nå å vise
\[
\sum_{cyc} (b+c)bc\sqrt{bc} \geq 2abc(a+b+c)
\]
La $x=\sqrt a$, $y=\sqrt b$ og $z = \sqrt{c}$. Ulikheten ovenfor blir da
\[
\sum_{cyc} (x^2+y^2)x^3y^3 \geq 2(x^2+y^2+z^2)x^2y^2z^2
\]
som følger av Muirhead. $\blacksquare$
\[
\sum_{cyc} \frac{b+c}{a}\sqrt{(a+b)(a+c)} \geq 4(a+b+c)
\]
AV CS på $\sqrt{(a+b)(a+c)}$ har vi $\frac{b+c}{a}\sqrt{(a+b)(a+c)}\geq \frac{b+c}{a}\sqrt{bc} + b+c$.
Det holder nå å vise
\[
\sum_{cyc} (b+c)bc\sqrt{bc} \geq 2abc(a+b+c)
\]
La $x=\sqrt a$, $y=\sqrt b$ og $z = \sqrt{c}$. Ulikheten ovenfor blir da
\[
\sum_{cyc} (x^2+y^2)x^3y^3 \geq 2(x^2+y^2+z^2)x^2y^2z^2
\]
som følger av Muirhead. $\blacksquare$