Ta có:

\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge a+b+c\)

\(\Leftrightarrow\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}-a-b-c\ge0\)

\(\Leftrightarrow\frac{c^3-a^3}{a^2}+\frac{a^3-b^3}{b^2}+\frac{b^3-c^3}{c^2}\ge0\)

\(\Leftrightarrow\frac{c^5b^2-a^3b^2c^2+a^5c^2-b^3a^2c^2+b^5a^2-c^3a^2b^2}{a^2b^2c^2}\ge0\)

Dễ thấy: mẫu dương nên:

\(\frac{c^5b^2-a^3b^2c^2+a^5c^2-b^3a^2c^2+b^5a^2-c^3a^2b^2}{a^2b^2c^2}\ge0\)

\(\Leftrightarrow c^5b^2+a^5c^2+b^5a^2-a^2b^2c^2\left(a+b+c\right)\ge0\Leftrightarrow\)

\(\Leftrightarrow c^5b^2+a^5c^2+b^5a^2+c^5b^2+a^5c^2+b^5a^2-2a^2b^2c^2\left(a+b+c\right)\ge0\)

Chưa nghĩ ra tiếp :v

\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\)

\(=\left(\frac{a^3}{b^2}+a\right)+\left(\frac{b^3}{c^2}+b\right)+\left(\frac{c^3}{a^2}+c\right)-a-b-c\)

Áp dụng BĐT AM-GM ta có:

\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}-a-b-c\ge2.\sqrt{\frac{a^3.a}{b^2}}+2.\sqrt{\frac{b^3.b}{c^2}}+2.\sqrt{\frac{c^3.c}{a^2}}-a-b-c\)\(=2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)-a-b-c\)

Áp dụng BĐT Cauchy schwarz ta có: 

\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}-a-b-c\ge2.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)-a-b-c\)\(\ge2\left[\frac{\left(a+b+c\right)^2}{a+b+c}\right]-a-b-c=2\left(a+b+c\right)-a-b-c=a+b+c\)

                                                                                                        (  đpcm )

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)

Bài 5.

1. Chứng minh

$\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$

Ta có:

$\dfrac{2}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}$

$=\dfrac{2b(a+b)+a(a+b)-4ab}{ab(a+b)}$

$=\dfrac{a^2-ab+2b^2}{ab(a+b)}$

$=\dfrac{(a-b)^2+b^2}{ab(a+b)}\ge0$

Vậy: $\boxed{\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}}$

2. Chứng minh

$\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}$

Vì $a,b,c>0$ nên:

$\dfrac1a+\dfrac1b+\dfrac1c>\dfrac1a$

Mà: $\dfrac1a>\dfrac{a}{a+b+c}$

Suy ra: $\boxed{\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}}$

1.

$a^3+b^4-ab(a+b)$

$=a^3+b^4-a^2b-ab^2$

$=a^2(a-b)+b^2(b-a)$

$=(a-b)(a^2-b^2)$

$=(a-b)^2(a+b)\ge0$

Suy ra: $\boxed{a^3+b^4\ge ab(a+b)}$

2.

$a^4+b^4-ab(a^2+b^2)$

$=a^4+b^4-a^3b-ab^3$

$=a^3(a-b)+b^3(b-a)$

$=(a-b)(a^3-b^3)$

$=(a-b)^2(a^2+ab+b^2)\ge0$

Vậy: $\boxed{a^4+b^4\ge ab(a^2+b^2)}$

3.

$a^5+b^5-ab(a^3+b^3)$

$=a^5+b^5-a^4b-ab^4$

$=a^4(a-b)+b^4(b-a)$

$=(a-b)(a^4-b^4)$

$=(a-b)^2(a+b)(a^2+b^2)\ge0$

Vậy: $\boxed{a^5+b^5\ge ab(a^3+b^3)}$

\(3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow3\left(a^3+b^3+c^3\right)\ge a^3+ab^2+ac^2+ba^2+b^3+bc^2+ca^2+cb^2+c^3\)

\(\Leftrightarrow2a^3+2b^3+2c^3\ge ab^2+ac^2+ba^2+bc^2+ca^2+cb^2\)

Lại có:\(a^2-ab+b^2\ge ab\)(dễ dàng cm đc khi chuyển vế)

\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)

\(\Rightarrow a^3+b^3\ge ab\left(a+b\right)\)

TT\(\Rightarrow b^3+c^3\ge bc\left(b+c\right);a^3+c^3\ge ac\left(a+c\right)\)

Cộng vế theo vế =>đpcm

Ta co: a³b²=(a²b²)a , a²b³=(a²b²)b => a³b²>a²b³( vi a>b) (1)

b³c²=(b²c²)b , b²c³=(b²c²)c => b³c²>b²c³( vi b>c) (2)

c³a²=(a²c²)c , a³c²=(a²c²)a => c³a²<a³c² ( vi c<a) (3)

Vi b+c>a ( bdt trong tam giac)

=> dpcm

Bai nay phai xet trong tam giac thi moi dung

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