Đặt (x3;y3;z3)=(a;b;c)(x,y,z>0)(x3;y3;z3)=(a;b;c)(x,y,z>0)

⇒xyz=1⇒xyz=1

Ta cần chứng minh

1x3+y3+1+1y3+z3+1+1z3+x3+1≤11x3+y3+1+1y3+z3+1+1z3+x3+1≤1

Áp dụng AM-GM, ta có: x3+y3+1=(x+y)(x2−xy+y2)+xyzx3+y3+1=(x+y)(x2−xy+y2)+xyz

≥(x+y)xy+xyz=xy(x+y+z)≥(x+y)xy+xyz=xy(x+y+z)

⇒1x3+y3+1≤1xy(x+y+z)⇒1x3+y3+1≤1xy(x+y+z)

Tương tự: 1y3+z3+1≤1yz(x+y+z)1y3+z3+1≤1yz(x+y+z)

1z3+x3+1≤1zx(x+y+z)1z3+x3+1≤1zx(x+y+z)

Cộng vế theo vế, ta được

....≤1x+y+z(1xy+1yz+1xz)=1x+y+z.x+y+zxyz=1xyz=1....≤1x+y+z(1xy+1yz+1xz)=1x+y+z.x+y+zxyz=1xyz=1

Vậy ta có đpcm

Đẳng thức xảy ra khi a=b=c=1

tại sao lời giải chẳng hiện ra thế

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

b)

Đề: Cho a, b, c > 0 và abc = ab + bc + ca. Chứng minh rằng: \(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}\le\frac{3}{16}\)

~ ~ ~ ~ ~

\(abc=ab+bc+ca\)

\(\Leftrightarrow1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Áp dụng BĐT \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\), ta có:

\(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}\)

\(\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{2\left(b+c\right)}+\frac{1}{2\left(a+b\right)}+\frac{1}{b+c}+\frac{1}{2\left(a+c\right)}+\frac{1}{a+b}\right)\)

\(=\frac{1}{4}\left[\frac{3}{2\left(a+c\right)}+\frac{3}{2\left(b+c\right)}+\frac{3}{2\left(a+b\right)}\right]\)

\(=\frac{3}{8}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{a+b}\right)\)

\(\le\frac{3}{32}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\frac{3}{16}\) (đpcm)

Dấu "=" xảy ra khi a = b = c 

Câu 2: \(\left(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\right)^2=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2+2\left(x^2+y^2+z^2\right)\)

\(=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2+6\)

Áp dụng bất đẳng thức AM - GM ta có :

\(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2\ge3\sqrt[3]{\left(\frac{xy}{z}\right)^2\left(\frac{yz}{x}\right)^2\left(\frac{xy}{y}\right)^2}=3\sqrt[3]{\frac{\left(xyz\right)^4}{\left(xyz\right)^2}}=3\)\(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge\sqrt{3+6}=3\left(dpcm\right)\)

tại sao lại suy ra đc \(3\sqrt[3]{\frac{\left(xyz\right)^4}{\left(xyz\right)^{^2}}}=3\) vậy cậu?

Ta có a+b+c>(a+b+c):1

=>a>1, b<1, c>1

=>.. dpcm

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=\frac{bc+ac+ab}{1}=bc+ac+ab\Rightarrow a+b+c>bc+ac+ab\)

\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(ab-a-b+1\right)\left(c-1\right)=abc-ac-bc+c-ab+a+b-1\)

\(=1-1+a+b+c-ac-bc-ab=a+b+c-\left(ac+bc+ab\right)\)

vì \(a+b+c>bc+ac+ab\)(chứng minh trên)\(\Rightarrow a+b+c-\left(bc+ac+ab\right)>0\)

\(\Rightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\)