Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH

A = -x² - 2xy - y² - 2y² + 10x + 10y + 4y - 25 + 7
= (-x² - 2xy - y² + 10x + 10y - 25) - 2y² + 4y + 7
= -(x² + 2xy + y² - 10x - 10y + 25) - (2y² - 4y - 7)
= -[(x+y)² - 10(x+y) + 25] - (2y² - 4y + 2 - 9)
= -(x + y - 5)² - 2(y² - 2y + 1) + 9
= -(x + y - 5)² - 2(y - 1)² + 9 ≤ 9
Dấu ''='' xảy ra <=> x + y - 5 = 0 và y -1 =0
<=> x + y = 5 và y = 1
<=> x = 4 và y = 1
Vậy max A = 9 <=> x = 4 và y = 1 .
- Mình chúc bạn học tốt nhé !

Áp dụng BĐT AM-GM ta có:

\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{c+1}{4}+\dfrac{b}{2}\ge3\sqrt[3]{\dfrac{a^3}{b\left(c+1\right)}\cdot\dfrac{c+1}{4}\cdot\dfrac{b}{2}}\)

\(=3\sqrt[3]{\dfrac{a^3}{4\cdot2}\cdot\dfrac{c+1}{c+1}\cdot\dfrac{b}{b}}=3\sqrt[3]{\dfrac{a^3}{8}}=\dfrac{3a}{2}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\dfrac{b^3}{c\left(a+1\right)}\ge\dfrac{3b}{2};\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3c}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT+\dfrac{a+b+c+3}{4}+\dfrac{a+b+c}{2}\ge\dfrac{3a+3b+3c}{2}\)

\(\Leftrightarrow VT+\dfrac{3\left(a+b+c\right)}{4}+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{2}\)

\(\Leftrightarrow VT+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{4}\). Mà theo AM-GM ta có:

\(a+b+c\ge3\sqrt[3]{abc}=3\)\(\Rightarrow VT+\dfrac{3}{4}\ge\dfrac{9}{4}\Rightarrow VT\ge\dfrac{3}{2}=VP\)

Đẳng thức xảy ra khi \(a=b=c=1\)

\(4x+8=y\left(3x-4\right)\Leftrightarrow3.4x+24=3y\left(3x-4\right)\)

\(4\left(3x-4\right)+40=3y\left(3x-4\right)\)

\(\left(3x-4\right)\left(3y-4\right)=40\)

Giải hệ nghiệm ngyên U 40

Hệ (I)

\(\left\{{}\begin{matrix}3x-4=8\\3y-4=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)

Hệ (II)

....

....

tự làm

Em bổ sung cho đủ đề nhé!

Hiếu Cao Huy bạn bổ xung chưa đủ

bổ xung đề (left{{}egin{matrix}x e0left(1 ight)\5x^2-5xy+y^2+dfrac{4}{x^2}=0left(2 ight)end{matrix} ight.) Tìm GTNN A=xy

Lời giải

(left(2 ight)Leftrightarrow5x^2-5xy+y^2+dfrac{4}{x^2}=0Leftrightarrowleft(x-y ight)^2+left(x-dfrac{1}{x} ight)^2+2-3xy=0)

(Leftrightarrowleft(x-y ight)^2+left(x-dfrac{1}{x} ight)^2=3xy-2)

(left{{}egin{matrix}left(x-y ight)^2ge0\left(x-dfrac{1}{x} ight)^2ge0end{matrix} ight.)(Rightarrow)VT (ge0Rightarrow3xy-2ge0Rightarrow xygedfrac{2}{3})

Kết luận

GTNN (A) =2/3

đạt được tại :x=y=+-1

Lỗi số học làm lại

\(\left(2\right)\Leftrightarrow\left(x-y\right)^2+4\left(x-\dfrac{1}{x}\right)^2+8-3xy=0\)

\(\left(2\right)\Leftrightarrow\left(x-y\right)^2+4\left(x-\dfrac{1}{x}\right)^2=3xy-8\)

VT>=0 => VP>=0 => xy>=8/3

GTNN (xy) =8/3

đạt được khi

x=y= +-1

Sửa phân số thứ nhất: \(\dfrac{a^2}{\sqrt{5a^2+32ab+b^2}}\rightarrow\dfrac{a^2}{\sqrt{5a^2+32ab+12b^2}}\)

Đề bài: \(P=\dfrac{a^2}{\sqrt{5a^2+32ab+12b^2}}+\dfrac{b^2}{\sqrt{5b^2+32bc+12c^2}}+\dfrac{c^2}{\sqrt{5c^2+32ac+12a^2}}\)

Lời giải

\(P=\dfrac{a^2}{\sqrt{5a^2+32ab+12b^2}}+\dfrac{b^2}{\sqrt{5b^2+32bc+12c^2}}+\dfrac{c^2}{\sqrt{5c^2+32ac+12a^2}}\)

\(\Leftrightarrow\dfrac{a^2}{\sqrt{5a^2+30ab+2ab+12b^2}}+\dfrac{b^2}{\sqrt{5b^2+30bc+2bc+12c^2}}+\dfrac{c^2}{\sqrt{5c^2+30ac+2ac+12a^2}}\)

\(\Leftrightarrow\dfrac{a^2}{\sqrt{5a\left(a+6b\right)+2b\left(a+6b\right)}}+\dfrac{b^2}{\sqrt{5b\left(b+6c\right)+2c\left(b+6c\right)}}+\dfrac{c^2}{\sqrt{5c\left(c+6a\right)+2a\left(c+6a\right)}}\)

\(\Leftrightarrow\dfrac{a^2}{\sqrt{\left(a+6b\right)\left(5a+2b\right)}}+\dfrac{b^2}{\sqrt{\left(b+6c\right)\left(5b+2c\right)}}+\dfrac{c^2}{\sqrt{\left(c+6a\right)\left(5c+2a\right)}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+6b\right)\left(5a+2b\right)}+\sqrt{\left(b+6c\right)\left(5b+2c\right)}+\sqrt{\left(c+6a\right)\left(5c+2a\right)}}\)

\(\Rightarrow VT\ge\dfrac{9}{\sqrt{\left(a+6b\right)\left(5a+2b\right)}+\sqrt{\left(b+6c\right)\left(5b+2c\right)}+\sqrt{\left(c+6a\right)\left(5c+2a\right)}}\) (1)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\left(a+6b\right)\left(5a+2b\right)}\le\dfrac{6a+8b}{2}\\\sqrt{\left(b+6c\right)\left(5b+2c\right)}\le\dfrac{6b+8c}{2}\\\sqrt{\left(c+6a\right)\left(5c+2a\right)}\le\dfrac{6c+8a}{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{\left(a+6b\right)\left(5a+2b\right)}+\sqrt{\left(b+6c\right)\left(5b+2c\right)}+\sqrt{\left(c+6a\right)\left(5c+2a\right)}\le\dfrac{14\left(a+b+c\right)}{2}=21\)

\(\Rightarrow\dfrac{9}{\sqrt{\left(a+6b\right)\left(5a+2b\right)}+\sqrt{\left(b+6c\right)\left(5b+2c\right)}+\sqrt{\left(c+6a\right)\left(5c+2a\right)}}\ge\dfrac{3}{7}\) (2)

Từ (1) và (2)

\(\Rightarrow VT\ge\dfrac{3}{7}\)

\(\Leftrightarrow\dfrac{a^2}{\sqrt{5a^2+32ab+12b^2}}+\dfrac{b^2}{\sqrt{5b^2+32bc+12c^2}}+\dfrac{c^2}{\sqrt{5c^2+32ac+12a^2}}\ge\dfrac{3}{7}\)

\(\Leftrightarrow P\ge\dfrac{3}{7}\)

Vậy \(P_{min}=\dfrac{3}{7}\)

Dấu " = " xảy ra khi \(a=b=c=1\)

lớp mấy 

Đặt \(a=x^2\left(a>=0\right)\)

pt trở thành \(a^2+\left(1-2m\right)a+m^2-1=0\)

\(\text{Δ}=\left(1-2m\right)^2-4\left(m^2-1\right)\)

\(=4m^2-4m+1-4m^2+4=-4m+5\)

a: Để pt vô nghiệm thì -4m+5<0

hay m>5/4

b: Để phương trình có hai nghiệm phân biệt thì -4m+5>0

hay m<5/4

c: Để pt có 4 nghiệm phân biệt thì 

\(\left\{{}\begin{matrix}m< \dfrac{5}{4}\\-2m+1>0\\m^2-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{5}{4}\\m< \dfrac{1}{2}\\\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m< -1\\\dfrac{1}{2}< m< 1\end{matrix}\right.\)

Ai tl giúp em với

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow3\ge ab+bc+ca\)

\(\Rightarrow\left\{{}\begin{matrix}3+a^2\ge\left(a+c\right)\left(a+b\right)\\3+b^2\ge\left(a+b\right)\left(b+c\right)\\3+c^2\ge\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{bc}{\sqrt{3+a^2}}\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}\\\dfrac{ca}{\sqrt{3+b^2}}\le\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\\\dfrac{ab}{\sqrt{3+c^2}}\le\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}+\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(\Leftrightarrow VT\le\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\) (1)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}\le\dfrac{\dfrac{bc}{a+c}+\dfrac{bc}{a+b}}{2}\\\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\dfrac{ab}{a+c}+\dfrac{ab}{b+c}}{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)+\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ab}{b+c}+\dfrac{ca}{b+c}\right)}{2}\)

\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\) (2)

Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(\Leftrightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

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

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)

\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (3)

Từ (1) , (2) , (3)

\(\Rightarrow VT\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(\Leftrightarrow\dfrac{bc}{\sqrt{a^2+3}}+\dfrac{ca}{\sqrt{b^2+3}}+\dfrac{ab}{\sqrt{c^2+3}}\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\) (đpcm)

Dấu " = " xảy ra khi \(a=b=c=1\)

Lời giải:

Áp dụng BĐT Cauchy cho $3$ số:

\(\left\{\begin{matrix} \frac{1}{a^3}+1+1\geq \frac{3}{a}\\ \frac{a^3}{b^3}+1+1\geq \frac{3a}{b}\\ b^3+1+1\geq 3b\end{matrix}\right.\Rightarrow \text{VT}\geq 3\text{VP}-6\)

Cũng áp dụng Cauchy:

\(\frac{1}{a}+\frac{a}{b}+b\geq 3\sqrt[3]{\frac{ab}{ab}}=3\Leftrightarrow \text{VP}\geq 3\)

\(\Rightarrow \text{VT}\geq 3\text{VP}-6\geq \text{VP}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=1\)

Ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

\(\Rightarrow ab+bc+ca=abc\)

Xét \(\dfrac{a^2}{a+bc}+\dfrac{b^2}{b+ca}+\dfrac{c^2}{c+ab}\)

\(\Leftrightarrow\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)

\(\Leftrightarrow\dfrac{a^3}{a^2+ab+bc+ca}+\dfrac{b^3}{b^2+ab+bc+ca}+\dfrac{c^3}{c^2+ab+bc+ca}\)

\(\Leftrightarrow\dfrac{a^3}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{b^3}{b\left(a+b\right)+c\left(a+b\right)}+\dfrac{c^3}{c\left(b+c\right)+a\left(b+c\right)}\)

\(\Leftrightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\\\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3b}{4}\\\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3b}{4}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{a+b+c}{2}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{a+b+c}{2}\)

\(\Rightarrow\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(b+c\right)\left(c+a\right)}\ge\dfrac{a+b+c}{4}\)

\(\Leftrightarrow\dfrac{a^2}{a+bc}+\dfrac{b^2}{b+ca}+\dfrac{c^2}{c+ab}\ge\dfrac{a+b+c}{4}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=3\)

p/s: bài này em nhớ em đã giải cho anh ròi mà ta =))

đài thế cách tui ngắn hơn nhiều