# Có bao nhiêu cách chứng minh cho bất đẳng thức Nesbit?

Trong topic nay chung ta se xem bat dang thuc

$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}\forall a,b,c>0$ co bao nhieu cach chung minh?

## 22 thoughts on “Có bao nhiêu cách chứng minh cho bất đẳng thức Nesbit?”

1. trungtuan says:

Cach 1. Cong them $1+1+1$ vao hai ve ta duoc bat dang thuc tuong duong $(a+b+c)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\geq \dfrac{9}{2}$ hay $[(a+b)+(b+c)+(c+a)]\left[\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right]\geq 9.$ Bat dang thuc nay dung! $\Box$

2. Phan Sy Quang says:

Cach 2.
Dat $P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$
$S=\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{a+b}$
$Q=\dfrac{c}{b+c}+\dfrac{a}{c+a}+\dfrac{b}{a+b}$
Ta co $Q+S=3$ va
$P+S=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\ge 3$
$P+Q=\dfrac{a+c}{b+c}+\dfrac{b+a}{c+a}+\dfrac{c+b}{a+b}\ge 3$
Suy ra $P\ge \dfrac{3}{2}.$

3. trungtuan says:

Cach 3.
Khong giam tong quat ta co the gia su $a+b+c=1$. Xet ham so $f(x)=\dfrac{x}{1-x}$ tren $I=(0,1)$, ta co $f''(x)=-\dfrac{2}{(x-1)^3}>0\forall x\in I$. Do do $f$ la ham so loi tren $I$, theo bat dang thuc Jensen ta co $f(a)+f(b)+f(c)\geq 3f(\dfrac{a+b+c}{3})=3f(\dfrac{1}{3})=\dfrac{3}{2}$. Nhu vay bat dang thuc duoc chung minh.$\Box$

4. Phan Sy Quang says:

Cach 4.
Dat $f(a,b,c)=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}$
Ta se chung minh:
$f(a,b,c)\ge f(\dfrac{a+b}{2},\dfrac{a+b}{2},c)$
That vay :
$f(a,b,c)- f(\dfrac{a+b}{2},\dfrac{a+b}{2},c)=\dfrac{(a-b)^2.(a+b+c)}{(2c+a+b)(b+c)(a+c)}\geq 0$
Vay ta co $f(a,b,c)\ge f(\dfrac{a+b}{2},\dfrac{a+b}{2},c)$
Tiep theo ta chung minh $f(\dfrac{a+b}{2},\dfrac{a+b}{2},c)\ge\dfrac{3}{2}$
That vay :Dat $t=\dfrac{a+b}{2}$
Bat dang thuc can chung minh tuong duong voi
$\dfrac{2t}{c+t}+\dfrac{c}{2t}\ge \dfrac{3}{2}$
Hay $\dfrac{(t-c)^2}{2t.(c+t)}\ge 0$
Hien nhien dung.
Quang

5. Phan Sy Quang says:

Cach 5.
Bien doi bat dang thuc ve dang SOS:
$\sum\dfrac{a}{b+c}-\dfrac{3}{2}=\sum\dfrac{(a-b)^2}{2(a+c).(b+c)}\ge 0$
BDT hien nhien dung vi $a,b,c>0$

6. Phan Sy Quang says:

Cach 6.
Bien doi ve dang SS.Gia su $c=\min(a,b,c)$
$\sum\dfrac{a}{b+c}-\dfrac{3}{2}=\dfrac{1}{(a+c)(b+c)}.(a-b)^2+\\ \dfrac{a+b+2c}{2(a+b)(b+c)(c+a)}.(a-c)(b-c)$
Hien nhien dung .

7. Phan Sy Quang says:

Cach 7.
Ap dung bat dang thuc Cauchy-Schwarz ta co :
$\sum\dfrac{a}{b+c}=\sum\dfrac{a^2}{ab+bc}$
$\ge \dfrac{(a+b+c)^2}{2(ab+bc+ca)}$
Theo bat dang thuc quen thuoc :
$(a+b+c)^2\ge 3(ab+bc+ca)$
Ta co dpcm .

8. Phan Sy Quang says:

Cach 8.
Bat dang thuc tuong duong voi
$\dfrac{\sum_{cyclic}a^3+[\sum_{sym}a^2.b]+3abc}{[\sum_{sym}a^2.b]+2abc}\ge \dfrac{3}{2}$
tuong duong voi :
$2.\sum_{cyclic}a^3\ge \sum_{sym}a^2.b$
Theo AM-GM:
$a^2.b\le \dfrac{a^3+a^3+b^3}{3}$
Cong ve theo ve 6 BDT tuong tu ta co dpcm.
done.

9. trungtuan says:

Cach 9.
Khong giam tong quat, gia su $a\geq b\geq c$ khi do $b+c\leq c+a\leq a+b$ hay $\dfrac{1}{b+c}\geq\dfrac{1}{c+a}\geq \dfrac{1}{a+b}$ (chu y la $a>0,b>0,c>0$) va boi bat dang thuc hoan vi ta co
$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq \dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{a+b}$(1) va $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq \dfrac{c}{b+c}+\dfrac{a}{c+a}+\dfrac{b}{a+b}$(2) .

Cong theo ve (1) voi (2) ta co dieu phai chung minh. $\Box$.

10. trungtuan says:

Cach 10.
Dat $b+c=2x,c+a=2y,a+b=2z$ suy ra $a=y+z-x,b=z+x-y,c=x+y-z$ va bat dang thuc da cho tro thanh $\dfrac{1}{2}\sum \dfrac{y+z-x}{x}\geq \dfrac{3}{2}$ hay $\sum_{\text{sym}}\dfrac{x}{y}\geq 6$, bat dang thuc cuoi cung nay dung theo AM-GM(chu y la $x>0,y>0,z>0\Box .$

11. trungtuan says:

Cach 11.
Khong giam tong quat , gia su $a+b+c=3$. Bang bien doi tuong duong ta chung minh duoc bat dang thuc sau $\dfrac{x}{3-x}\geq\dfrac{3}{4}(x-1)+\dfrac{1}{2}(0(*). Lan luot thay $x$ boi $a,b,c$ trong (*) chung ta duoc ba bat dang thuc, cong theo ve ba bat dang thuc nay chung ta duoc dieu phai chung minh.$\Box$

12. Tuan says:

Cach 12.
Dat $\dfrac{a}{b+c}=x,\dfrac{b}{c+a}=y,\dfrac{c}{a+b}=z$ thi $xy+yz+zx+2xyz=1$ va chung ta can chung minh $x+y+z\geq \dfrac{3}{2}$. Gia su nguoc lai $x+y+z< \dfrac{3}{2}$, khi do $1=xy+yz+zx+2xyz\leq \dfrac{(x+y+z)^2}{3}+2\left(\dfrac{x+y+z}{3}\right)^3<1$, vo li! Suy ra dieu phai chung minh.$\Box$

13. Phan Sy Quang says:

Cach 13.
Chuan hóa cho $\ a+b+c=1$

Ta có:
$\dfrac{a}{b+c}+\dfrac{9a(b+c)}{4}\geq 3a$ tương tu voi các bieu thuc khác sau đó áp dung thêm bdt:
$(a+b+c)^2\geq 3(ab+bc+ca)$ ta có đpcm . $\Box$

14. Phan Sy Quang says:

Cach 14.
Ta chung minh nhan xet sau
$\dfrac{a}{b+c}\geq \dfrac{8a-b-c}{4(a+b+c)}$

That vay,no tuong duong voi:$(2a-b-c)^2\ge 0$
hien nhien dung.Cong ve theo ve cac bat dang thuc tuong tu ta co ngay dpcm.
done.

15. Phan Sy Quang says:

Cach 15.
Gia su $a\ge b\ge c$.Suy ra:
$\dfrac{1}{b+c}\ge \dfrac{1}{c+a}\ge \dfrac{1}{a+b}$

Theo bat dang thuc Chebyshev,ta co :

$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \ge$

$\dfrac{1}{3}.(a+b+c)(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}) =$

$\dfrac{1}{6}[(a+b)+(b+c)+(c+a)][\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}]\ge\dfrac{9}{6}$ ( theo AM-GM)
ta co dieu can phai chung minh.

16. Phan Sy Quang says:

Cach 16.
Theo Schwarz ta co :
$\sum_{cyclic}\dfrac{a}{b+c}=\sum_{cyclic}\dfrac{a^4}{a^3(b+c)}$
$\ge\dfrac{(a^2+b^2+c^2)^2}{\sum_{sym}a^3.b}\ge \dfrac{3}{2}$
Ta can chung minh :$2(a^2+b^2+c^2)^2\ge 3\sum_{sym}a^3.b$
Day la 1 bat dang thuc rat noi tieng cua Vasile Cirtoaje.

17. nguyenducminh says:

Cach 17.
$\dfrac{a(b+c)}{b+c} + \dfrac{a^{2}}{b+c} + \dfrac{b+c}{4} \geq 2a$

suy ra
$\dfrac{a(a+b+c)}{b+c} + \dfrac{b+c}{4} \geq 2a$
tuong tu
$\dfrac{b(a+b+c)}{a+c} + \dfrac{a+c}{4} \geq 2b$
va

$\dfrac{c(a+b+c)}{a+b} + \dfrac{b+a}{4} \geq 2c$
cong vao :
$(a+b+c)( \dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{b+a} ) \geq \dfrac{3}{2}(a+b+c)$

18. trungtuan says:

Cach 18(Thu luom tren Internet).
Dat $x=\dfrac{a}{b+c},y=\dfrac{b}{c+a},z=\dfrac{c}{a+b}$ va xet ham so $f(t)=\dfrac{t}{t+1}(t>0)$. Ta co $f'(t)=\dfrac{1}{(t+1)^2}>0\forall t>0$ va $f''(t)=-\dfrac{2}{(t+1)^3}<0\forall t\in (0,\infty)$. Theo bat dang thuc Jensen ta co $f(\dfrac{1}{2})=\dfrac{1}{3}=\dfrac{f(x)+f(y)+f(z)}{3}\leq f\left(\dfrac{x+y+z}{3}\right)$. Nhung vi $f$ la tang ngat tren $(0,\infty)$ nen $\dfrac{1}{2}\leq \dfrac{x+y+z}{3}$. Suy ra dieu phai chung minh.$\Box$

19. trungtuan says:

Cach 19.(Tren internet)
Se la du neu ta chung minh duoc $\dfrac{a}{b+c}\geq \dfrac{3a^{3/2}}{2(a^{3/2}+b^{3/2}+c^{3/2})}$ hay $2(a^{3/2}+b^{3/2}+c^{3/2})\geq 3a^{1/2}(b+c)$(*).

Theo AM-GM ta co
$a^{3/2}+b^{3/2}+b^{3/2}\geq 3a^{1/2}b$ va
$a^{3/2}+c^{3/2}+c^{3/2}\geq 3a^{1/2}c$ .
Cong theo ve hai bat dang thuc cuoi nay ta duoc (*).$\Box$

20. ductrung says:

Zui qua, post thu 1 cai:
$\sum_{cycl}\dfrac{a}{b+c}=\dfrac{1}{2}\sum_{cycl}\dfrac{2a}{b+c}$

$=\dfrac{1}{2}\sum_{cycl}\left(\dfrac{(c+a)+(a+b)}{b+c}-1\right)$

$\ge\dfrac{1}{2}(6-3)=\dfrac{3}{2}$.

21. ductrung says:

$\sum_{cycl}\left(\dfrac{a}{b+c}-\dfrac{1}{2}\right)$

$=\dfrac{1}{2(a+b)(b+c)(c+a)}\sum_{cycl} (2a+b+c)(a-b)(a-c)$
which is true by Vornicu-Schur lemma.