Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!

Lời giải:

Từ \(xy+yz+xz=xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Đặt \((a,b,c)=\left(\frac{1}{x}; \frac{1}{y}; \frac{1}{z}\right)\Rightarrow a+b+c=1\)

BĐT cần chứng minh trở thành:

\(P=\frac{c^3}{(a+1)(b+1)}+\frac{a^3}{(b+1)(c+1)}+\frac{b^3}{(c+1)(a+1)}\geq \frac{1}{16}(*)\)

Thật vậy, áp dụng BĐT Cauchy ta có:

\(\frac{c^3}{(a+1)(b+1)}+\frac{a+1}{64}+\frac{b+1}{64}\geq 3\sqrt[3]{\frac{c^3}{64^2}}=\frac{3c}{16}\)

\(\frac{a^3}{(b+1)(c+1)}+\frac{b+1}{64}+\frac{c+1}{64}\geq 3\sqrt[3]{\frac{a^3}{64^2}}=\frac{3a}{16}\)

\(\frac{b^3}{(c+1)(a+1)}+\frac{c+1}{64}+\frac{a+1}{64}\geq 3\sqrt[3]{\frac{b^3}{64^2}}=\frac{3b}{16}\)

Cộng theo vế các BĐT trên và rút gọn :

\(\Rightarrow P+\frac{a+b+c+3}{32}\geq \frac{3(a+b+c)}{16}\)

\(\Leftrightarrow P+\frac{4}{32}\geq \frac{3}{16}\Leftrightarrow P\geq \frac{1}{16}\)

Vậy \((*)\) được chứng minh. Bài toán hoàn tất.

Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=3\)

\(BĐT\Leftrightarrow\dfrac{x}{y^3}+\dfrac{y}{z^3}+\dfrac{z}{x^3}\ge x+y+z\)

Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Rightarrow abc\ge1\)

\(BĐT\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}=\dfrac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\)

Ta có \(abc\ge1\)

\(\Rightarrow\left\{{}\begin{matrix}bc\ge\dfrac{1}{a}\\ab\ge\dfrac{1}{c}\\ac\ge\dfrac{1}{b}\end{matrix}\right.\Rightarrow bc+ac+ab\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)

\(\Leftrightarrow\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\ge0\)

\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{\left(1+y\right)\left(1+z\right).64}}=\dfrac{3x}{4}\)

\(\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3y}{4}\)

\(\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{8}+\dfrac{1+y}{8}\ge\dfrac{3z}{4}\)

\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{x+y+z}{2}-\dfrac{3}{4}\ge\dfrac{3\sqrt[3]{xyz}}{2}-\dfrac{3}{4}=\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}\left(đpcm\right)\)

(bài này chắc thiếu đk xyz=1 ?nên mình bổ sung xyz=1)

( xyz=3)

Áp dụng BDDT AM-GM:

Ta có: \(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{\left(1+y\right)\left(1+z\right).8.8}}=3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)

Chứng minh tương tự ta có:

\(\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3y}{4}\)

\(\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{8}+\dfrac{1+y}{8}\ge\dfrac{3z}{4}\)

Cộng từng vế ta được:

\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{3+x+y+z}{4}\ge\dfrac{3\left(x+y+z\right)}{4}\)

\(\Leftrightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{3x+3y+3z-3-x-y-z}{4}=\dfrac{2\left(x+y+z\right)-3}{4}\)

\(\Leftrightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{2.\sqrt[3]{xyz}-3}{4}=\dfrac{2.3-3}{4}=\dfrac{3}{4}\left(đfcm\right)\)

Gọi cái thiệt gớm đó là P

Ta có:

\(xy+yz+zx=xyz\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

Ta có:

\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64y}+\dfrac{1+y}{64x}\ge3\sqrt[3]{\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}.\dfrac{1+x}{64y}.\dfrac{1+y}{64x}}=\dfrac{3}{16z}\)

\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{64x}-\dfrac{1}{64y}-\dfrac{1}{32}\left(1\right)\)

Tương tự ta cũng có:

\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{64y}-\dfrac{1}{64z}-\dfrac{1}{32}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{64z}-\dfrac{1}{64x}-\dfrac{1}{32}\left(3\right)\end{matrix}\right.\)

Từ (1), (2), (3) ta được

\(P\ge\dfrac{3}{16}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)

\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)

Dấu = xảy ra khi \(x=y=z=3\)

Đặt cái ban đầu là P

Ta có: \(xy+yz+zx=xyz\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

Ta lại có:

\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)

\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)

Tương tự ta có:

\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)

Từ (1), (2), (3) ta có:

\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)

\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)

Dấu = xảy ra khi \(x=y=z=3\)

Hầy mình không nghĩ lớp 7 đã phải làm những bài biến đổi như thế này. Cái này phù hợp với lớp 8-9 hơn.

1.

Đặt $x^2-y^2=a; y^2-z^2=b; z^2-x^2=c$. 

Khi đó: $a+b+c=0\Rightarrow a+b=-c$

$\text{VT}=a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3$

$=(-c)^3-3ab(-c)+c^3=3abc$

$=3(x^2-y^2)(y^2-z^2)(z^2-x^2)$

$=3(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)$

$=3(x-y)(y-z)(z-x)(x+y)(y+z)(x+z)$

$=3.4(x-y)(y-z)(z-x)=12(x-y)(y-z)(z-x)$

Ta có đpcm.

Bài 2:

Áp dụng kết quả của bài 1:

Mẫu:

$(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3=3(x-y)(y-z)(z-x)(x+y)(y+z)(z+x)=3(x-y)(y-z)(z-x)(1)$

Tử: 

Đặt $x-y=a; y-z=b; z-x=c$ thì $a+b+c=0$

$(x-y)^3+(y-z)^3+(z-x)^3=a^3+b^3+c^3$

$=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3=3abc$

$=3(x-y)(y-z)(z-x)(2)$

Từ $(1);(2)$ suy ra \(\frac{(x-y)^3+(y-z)^3+(z-x)^3}{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}=1\)

Bài 3:

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

Do đó:

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

Ta có đpcm.

Chứng minh $x+2y+z\geq 4(1-x)(1-y)(1-z)$ - Bất đẳng thức và cực trị - Diễn đàn Toán học

Ta có: \(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge\dfrac{3x}{4}\)

\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\dfrac{6x-y-z-2}{8}\left(1\right)\)

Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\dfrac{6y-z-x-2}{8}\left(2\right)\\\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{6z-x-y-2}{8}\left(3\right)\end{matrix}\right.\)

Từ (1), (2), (3)

\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{6x-y-z-2}{8}+\dfrac{6y-z-x-2}{8}+\dfrac{6z-x-y-2}{8}\)

\(=\dfrac{1}{2}\left(x+y+z\right)-\dfrac{3}{4}\ge\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}\)

fix đề: CMR:\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}\)

Áp dụng AM-GM có:

\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{8\cdot8\cdot\left(1+y\right)\left(1+z\right)}}=3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)

Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3y}{4}\\\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}+\dfrac{1+y}{8}+\dfrac{1+x}{8}\ge\dfrac{3z}{4}\end{matrix}\right.\)

Cộng theo về các BĐT trên ta được:

\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}+\dfrac{3+x+y+z}{4}\ge\dfrac{3\left(x+y+z\right)}{4}\)

\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}\ge\dfrac{3x+3y+3z-x-y-z-3}{4}=\dfrac{2\left(x+y+z\right)-3}{4}\)

\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}\ge\dfrac{2\cdot3\sqrt[3]{xyz}-3}{4}=\dfrac{2\cdot3-3}{4}=\dfrac{3}{4}\)

-> Đpcm

Dấu ''='' xảy ra khi x = y = z = 1

Hóng với. T cũng định up bài này