a2+b2+c2=1a2+b2+c2=1

|a|;|b|;|c|≤1|a|;|b|;|c|≤1

−1≤a;b;c≤1−1≤a;b;c≤1

(a+1)(b+1)(c+1)≥0(a+1)(b+1)(c+1)≥0

ab+bc+ac+a+b+c+1+abc≥0(1)ab+bc+ac+a+b+c+1+abc≥0(1)

Mặt khác ta có :

(1+a+b+c)2≥0(1+a+b+c)2≥0

a2+b2+c2+2(ab+bc+ac)+2(a+b+c)+1≥0a2+b2+c2+2(ab+bc+ac)+2(a+b+c)+1≥0

2(a+b+c+ab+bc+ac+1)≥02(a+b+c+ab+bc+ac+1)≥0

(a+b+c+ab+bc+ac+1)≥0(2)(a+b+c+ab+bc+ac+1)≥0(2)

trong nâng cao và phát triển có bài này thật đấy

post ít một thôi

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(VT=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac+ab+bc+ac+a^2+b^2+c^2}+\dfrac{7}{ab+bc+ac}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\)

Áp dụng bất đẳng thức AM-GM cho 2 số dương:

\(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1^2}{3}=\dfrac{1}{3}\)

Ta có: \(\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)

Áp dụng BĐT Cauchy-Schwarz ta có

BT\(\ge\)\(\frac{\left(1+1+1\right)^2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}=\frac{9}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}\)

\(=\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}+\frac{7}{ab+bc+ac}\)

\(\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}+\frac{7}{ab+bc+ac}\)\(=1+\frac{7}{ab+bc+ac}\)

Ta lại có ab+bc+ac =< (a+b+c)^2/3 =3

\(\Rightarrow BT\ge1+\frac{7}{3}=\frac{10}{3}\)

Vậy GTNN là \(\frac{10}{3}\)khi a=b=c=1

Cô-si Schwarzt dạng Engel là đc

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: $\boxed{a^5+b^5\ge ab(a^3+b^3)}$