Točno
8. prosinca 2017. 23:54 (6 godine, 7 mjeseci)
If
![a, b, c](/media/m/9/e/9/9e9dfe78930065fbe5a777e9b07c27c4.png)
are three positive real numbers such that
![ab+bc+ca = 1](/media/m/3/1/5/315be27fca0bcae6050ae5a6015fe406.png)
, prove that
%V0
If $a, b, c$ are three positive real numbers such that $ab+bc+ca = 1$, prove that $$\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}.$$
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Promatrajmo funkciju
![f(x)=x^{\frac{1}{3}}](/media/m/5/4/0/540a8b19b6ddb87ba7ea4af2c3bcf0a4.png)
odakle proizlazi
![f''(x)=-\frac{2}{9}x^{-\frac{5}{3}}\leq 0](/media/m/4/c/6/4c6f2e77910db482937b1b07376858e8.png)
dakle možemo po Jensenovj nejednakosti zaključiti da vrijedi
![\sum_{cyc} f\bigg(\frac{1}{a}+6b\bigg) \leq 3 f\bigg(\frac{1}{3}\big(\sum_{cyc} \frac{1}{a} + \sum_{cyc} 6a \big)\bigg)](/media/m/1/c/5/1c5500af98b2b4b07a6c341b22389695.png)
dakle ostaje nam dokazati da je
![\iff 9a^2b^2c^2 +54\sum_{cyc}a^4b^3c^3\leq 1](/media/m/9/6/a/96abd59e0a06b3af52c3c3082d598487.png)
Što dovršavama vrlo jednostavno sa dvije nejednakosti
![3\sqrt[3]{a^2b^2c^2} \leq ab+bc+ca =1 \implies a^2b^2c^2\leq\frac{1}{27}](/media/m/6/e/2/6e2b32b454cd22c308348efa4f1585fe.png)
![abc(a+b+c)\leq\frac{(ab+bc+ca)^2}{3}=\frac{1}{3}](/media/m/9/c/5/9c5fec13e6eadc40cf594dad9837c502.png)
Dakle
%V0 Promatrajmo funkciju $f(x)=x^{\frac{1}{3}}$ odakle proizlazi $f''(x)=-\frac{2}{9}x^{-\frac{5}{3}}\leq 0$ dakle možemo po Jensenovj nejednakosti zaključiti da vrijedi
$$\sum_{cyc} f\bigg(\frac{1}{a}+6b\bigg) \leq 3 f\bigg(\frac{1}{3}\big(\sum_{cyc} \frac{1}{a} + \sum_{cyc} 6a \big)\bigg)$$ dakle ostaje nam dokazati da je $$\sqrt[3]{9\frac{1}{abc} +54\sum_{cyc}a}\leq \frac{1}{abc}$$ $$\iff 9a^2b^2c^2 +54\sum_{cyc}a^4b^3c^3\leq 1$$
Što dovršavama vrlo jednostavno sa dvije nejednakosti $$ 3\sqrt[3]{a^2b^2c^2} \leq ab+bc+ca =1 \implies a^2b^2c^2\leq\frac{1}{27}$$
$$abc(a+b+c)\leq\frac{(ab+bc+ca)^2}{3}=\frac{1}{3}$$
Dakle $$9a^2b^2c^2 +54\sum_{cyc}a^4b^3c^3=a^2b^2c^2 \big( 9 + 54 abc ( a + b+ c) \big) \leq \frac{1}{27}\big( 9 +54 \frac{1}{3}\big)=1$$
10. prosinca 2017. 01:59 | Lugo | Točno |