Točno
22. listopada 2020. 19:06 (3 godine, 8 mjeseci)
Neka su $a$, $b$, $c$ $ \in \mathbb{R}$ takvi da vrijedi $0 \leq a$, $b$, $c \leq 1$. Dokaži nejednakost: $$a^2+b^2+c^2 \leq a^2b+b^2c+c^2a+1.$$
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![\textbf{Prvo rješenje}](/media/m/7/2/e/72eae00964f4f0a5b9d230a19220e595.png)
Uvedimo supstituciju
![x=1-a,y=1-b,z=1-c](/media/m/5/d/5/5d5f6dbccf1a626dfd159a812666cfb1.png)
. Vrijedi
![0 \leq x,y,z \leq 1](/media/m/1/4/f/14f78fa56e967320821275d4bd37906c.png)
i tvrdnja postaje
![(1-x)^2 + (1-y)^2+(1-z)^2 \leq (1-x)^2(1-y)+(1-y)^2(1-z)+(1-z)^2(1-x)+ (1-x)(1-y)(1-z)](/media/m/2/3/f/23fab0b2645fffb31a5851125f5f263b.png)
.
![\iff x+y+z+x^2y+y^2z+z^2x\leq1+2xy+2yz+2xz](/media/m/5/1/1/511d89fad422a65a0fff74b73c13709f.png)
Primijetimo da vrijedi
![(1-x)(1-y)(1-z)\geq0](/media/m/1/3/4/1340c743481a52f28e009d0e8d776e34.png)
što implicira
![1+xy+yz+xz\geq x+y+z+xyz](/media/m/3/9/1/39130216a157dc22f32d7e78ea19e6c3.png)
. Ako toj nejednakosti dodamo
![xy+yz+xz](/media/m/9/4/0/9405dd972a6acba263b3b2e70490bcab.png)
na svaku stranu, dobivamo da vrijedi
![1+2xy+2yz+2xz\geq x+y+z+xy+yz+xz+xyz](/media/m/2/c/e/2ced6a9cfa457917e8bbbe60f7df7f39.png)
.
Ostaje nam pokazati da vrijedi
![xy+yz+xz+xyz\geq x^2y+y^2z+z^2x](/media/m/8/9/9/899e8b6aa18946c58467562bdde8af0c.png)
,
ali to je očito zbog
![xy\geq x^2y,\ yz\geq y^2z,\ xz\geq z^2x](/media/m/0/7/7/077391d5643c10394e76fc29824e10bb.png)
.
![\textbf{Drugo rješenje}](/media/m/f/b/b/fbbc36f54e4a88e1b3ffbd5755e51534.png)
Vidimo da vrijedi
![(1-a^2)(1-b^2)(1-c^2)\geq 0](/media/m/a/7/b/a7bdafae26de5c1f5b65062a719b5bca.png)
iz čega slijedi
![1+a^2b^2+b^2c^2+a^2c^2 \geq a^2+b^2+c^2+a^2b^2c^2](/media/m/c/8/5/c855b506d098039e053a0ee9a284ef1c.png)
.
Dovoljno je pokazati
![1+a^2b+b^2c+c^2a \geq 1+a^2b^2+b^2c^2+a^2c^2](/media/m/1/f/f/1ffac398e43cbbb4233a94eec5434a9a.png)
što je očito zbog
![a^2b \geq a^2b^2,\ b^2c \geq b^2c^2,\ c^2a \geq a^2c^2](/media/m/4/3/6/4369b8df771b5fe94d2643f8796bdee3.png)
.
%V0
$\textbf{Prvo rješenje}$
Uvedimo supstituciju $x=1-a,y=1-b,z=1-c$. Vrijedi $0 \leq x,y,z \leq 1$ i tvrdnja postaje
$$(1-x)^2 + (1-y)^2+(1-z)^2 \leq (1-x)^2(1-y)+(1-y)^2(1-z)+(1-z)^2(1-x)+ (1-x)(1-y)(1-z)$$.
$$\iff x+y+z+x^2y+y^2z+z^2x\leq1+2xy+2yz+2xz$$
Primijetimo da vrijedi $(1-x)(1-y)(1-z)\geq0$ što implicira $1+xy+yz+xz\geq x+y+z+xyz$. Ako toj nejednakosti dodamo $xy+yz+xz$ na svaku stranu, dobivamo da vrijedi
$$1+2xy+2yz+2xz\geq x+y+z+xy+yz+xz+xyz$$.
Ostaje nam pokazati da vrijedi
$$xy+yz+xz+xyz\geq x^2y+y^2z+z^2x$$,
ali to je očito zbog $xy\geq x^2y,\ yz\geq y^2z,\ xz\geq z^2x$.
$\textbf{Drugo rješenje}$
Vidimo da vrijedi $(1-a^2)(1-b^2)(1-c^2)\geq 0$ iz čega slijedi
$$1+a^2b^2+b^2c^2+a^2c^2 \geq a^2+b^2+c^2+a^2b^2c^2$$.
Dovoljno je pokazati $1+a^2b+b^2c+c^2a \geq 1+a^2b^2+b^2c^2+a^2c^2$ što je očito zbog $a^2b \geq a^2b^2,\ b^2c \geq b^2c^2,\ c^2a \geq a^2c^2$.
22. listopada 2020. 20:06 | 11235 | Točno |