Chương I - Căn bậc hai. Căn bậc ba

Thôi đang rảnh, giúp bạn bài này luôn vậy!!

Giải:

Ta có:

\(VT=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\left(\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}+\dfrac{a^2}{a+b}\right)=A+B\)

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

\(\ge\dfrac{1}{2}3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}3\sqrt[3]{\dfrac{1}{a+b}\dfrac{1}{b+c}\dfrac{1}{c+a}}=\dfrac{9}{2}\)

\(\Rightarrow A\ge\dfrac{3}{2}\)

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

\(\Leftrightarrow1\le B.2\Leftrightarrow B\ge\dfrac{1}{2}\)

Từ đó ta có: \(VT\ge\dfrac{3}{2}+\dfrac{1}{2}=2=VP\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

\(\dfrac{a+b^2}{b+c}+\dfrac{b+c^2}{c+a}+\dfrac{c+a^2}{a+b}\ge2\)

\(\Leftrightarrow\dfrac{a\left(a+b+c\right)+b^2}{b+c}+\dfrac{b\left(a+b+c\right)+c^2}{c+a}+\dfrac{c\left(a+b+c\right)+a^2}{a+b}\ge2\)

\(\Leftrightarrow\dfrac{a^2+ab+ac+b^2}{b+c}+\dfrac{ab+b^2+bc+c^2}{c+a}+\dfrac{ca+bc+c^2+a^2}{a+b}\ge2\)

\(\Leftrightarrow\dfrac{a^2+b^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+c^2+b\left(c+a\right)}{c+a}+\dfrac{c^2+a^2+c\left(a+b\right)}{a+b}\ge2\)

\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}+1\ge2\)

\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}\ge1\)

\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge1\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

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

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

Áp dụng bất đẳng thức Mincopski

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

\(\Rightarrow\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2\ge2\)

\(\Rightarrow\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2}\ge1\)

\(\Rightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge1\)

\(\Leftrightarrow\dfrac{a+b^2}{b+c}+\dfrac{b+c^2}{c+a}+\dfrac{c+a^2}{a+b}\ge2\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Đặt \(\left(a+1;b+1;c+1\right)\rightarrow\left(x;y;z\right)\).Giả thiết trở thành:\(xyz=x+y+z\) và cần tìm max của \(P=\sum\dfrac{x}{x^2+1}\)

Ta có: \(P=\sum\dfrac{x}{x^2+1}=\sum\dfrac{xyz}{x\left(x+y+z\right)+yz}=xyz.\sum\dfrac{1}{\left(x+y\right)\left(x+z\right)}\)

\(=\dfrac{2xyz\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

Do \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\) nên \(P\le\dfrac{2xyz}{\dfrac{8}{9}\left(xy+yz+xz\right)}=\dfrac{9}{4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}\)(*)

Mặt khác , từ giả thiết ta có : \(1=\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\le\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\)( theo AM-GM)

\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\sqrt{3}\)

Kết hợp với (*) , ta suy ra \(P\le\dfrac{9}{4\sqrt{3}}=\dfrac{3\sqrt{3}}{4}\)

Dấu = xảy ra khi \(x=y=z=\sqrt{3}\) hay \(a=b=c=\sqrt{3}-1\)

P/s: Chứng minh \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

khai triển ra ta có: \(\sum ab\left(a+b\right)\ge6abc\)hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)( đúng)

Không mất tính tổng quát giả sử: \(x\ge y\ge z\ge t>0\)

\(\Rightarrow1=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{1}{t^2}\le\dfrac{4}{t^2}\)

\(\Leftrightarrow t^2\le4\)

\(\Leftrightarrow0< t\le2\)

\(\Rightarrow t=\left\{1,2\right\}\)

Cứ vậy sẽ giải được bài toán

Wlog x\ge y\ge z\ge t

\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\)

\(=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{34}{ab}+\dfrac{17}{8}ab-\dfrac{1}{8}ab\)

\(\ge2.\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{34}{ab}.\dfrac{17}{8}ab}-\dfrac{1}{8}.\dfrac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow A\ge2.\dfrac{4}{\left(a+b\right)^2}+2.\dfrac{17}{2}-\dfrac{1}{8}.\dfrac{4^2}{4}\ge2.\dfrac{4}{4^2}+17-\dfrac{1}{2}\)

\(\Leftrightarrow A\ge\dfrac{1}{2}+17-\dfrac{1}{2}=17\)

Dấu "=" <=> a = b = 2

a/ Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\)

\(\Rightarrow10ab=3\left(a^2+b^2\right)\)

\(\Leftrightarrow\left(3a-b\right)\left(3b-a\right)=0\)

b/ Nó có phải là phương trình đâu

b/ \(\left(x^2+1\right)\left(y^2+2\right)\left(z^2+8\right)\ge2x.2\sqrt{2}y.2\sqrt{8}z=32xyz\)

@Cold Wind

Ta có:

\(\left(\sqrt{8-x}+\sqrt{8-y}+\sqrt{8-z}\right)^2\le3\left(24-x-y-z\right)\)

\(\le3\left(24-\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}\right)=36\)

\(\Rightarrow\sqrt{8-x}+\sqrt{8-y}+\sqrt{8-z}^2\le6\)

Dấu = xảy ra khi \(x=y=z=2\)

khó và rộng quá chịu

Lời giải:

Với những bài như này em chỉ cần nắm rõ điểm rơi rồi phân tích hợp lý để áp dụng những BĐT quen thuộc là được.

Ta có:

\(P=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}=\frac{3(a+b)}{4\sqrt{ab}}+\frac{a+b}{\sqrt{4ab}}+\frac{\sqrt{ab}}{a+b}\)

Áp dụng BĐT AM-GM ta có:

\(a+b\geq 2\sqrt{ab}\Rightarrow 3(a+b)\geq 6\sqrt{ab}\Rightarrow \frac{3(a+b)}{4\sqrt{ab}}\geq \frac{6\sqrt{ab}}{4\sqrt{ab}}=\frac{3}{2}\)

Và:

\(\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\geq 2\sqrt{\frac{1}{4}}=1\)

Do đó:

\(P=\frac{3(a+b)}{4\sqrt{ab}}+\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\geq \frac{3}{2}+1=\frac{5}{2}\)

Vậy \(P_{\min}=\frac{5}{2}\)

Dấu bằng xảy ra khi \(a=b\)

Cold Wind không cần kiểu mò mẫn (điểm rơi ) .

\(t=\dfrac{a+b}{\sqrt{ab}}\) quá đơn giản nhận ra \(t\ge2\)

\(P\left(t\right)=t+\dfrac{1}{t}=\dfrac{t^2+1}{t}=m\Leftrightarrow\left\{{}\begin{matrix}t^2-mt+1=0\\t\ge2\end{matrix}\right.\)\(\begin{matrix}\left(1\right)\\\left(2\right)\end{matrix}\)

(1)có nghiệm<=> :\(\left\{{}\begin{matrix}m\in\left(-vc;-2\right)U\left(2;vc\right)\\t=\dfrac{m\pm\sqrt{m^2-4}}{2}\end{matrix}\right.\)

\(t\ge2\Leftrightarrow\dfrac{m+\sqrt{m^2-4}}{2}\ge2\Leftrightarrow\sqrt{m^2-4}\ge4-m\)

m>4 luôn đúng

xét \(m\le4\) \(\Leftrightarrow m^2-4\ge16-8m+m^2\Leftrightarrow m\ge\dfrac{20}{8}=\dfrac{5}{2}\)

\(\Rightarrow P_{min}=\dfrac{5}{2}\) khi t =2 <=> a=b>0

\(\sqrt{2x+1}+\dfrac{2x-1}{x+3}-\left(2x-1\right)\sqrt{x^2+4}-\sqrt{2}=0\)

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

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

\(\Leftrightarrow x=\dfrac{1}{2}\)

PS: Phần trong ngoặc chứng minh vô nghiệm cũng không khó b tự làm nốt nhé.

Lời giải:

Đặt \(A=(a+1)(b+1)(c+1)\)

\(6A=(a+1)(b+b+2)(c+c+c+3)\)

Áp dụng BĐT AM-GM ta có:

\(6A\geq 2\sqrt{ab}.3\sqrt[3]{2b^2}.4\sqrt[4]{3c^3}\)

\(\Leftrightarrow 6A\geq 24\sqrt{a}.\sqrt[3]{2b^2}.\sqrt[4]{3c^3}=24\sqrt[12]{a^6.16b^8.27c^9}\)

\(\Leftrightarrow A\geq 4\sqrt[12]{432a^6b^8c^9}\) (1)

Lại có:

\(abc=ab(6-a-b)=\frac{2}{9}.3a.\frac{3}{2}b(6-a-b)\)

\(\leq \frac{2}{9}.\left(\frac{3a+\frac{3}{2}b+6-a-b}{3}\right)^3\) (BĐT AM-GM ngược dấu)

\(\Leftrightarrow abc\leq \frac{2}{9}\left(\frac{6+2a+\frac{b}{2}}{3}\right)^3\leq \frac{2}{9}\left(\frac{6+2+1}{3}\right)^3\)

\(\Leftrightarrow abc\leq 6\) (2)

Từ (1); (2) suy ra \(A\geq 4\sqrt[12]{2.(abc)^3.a^6b^8c^9}\geq 4\sqrt[12]{a^3b.a^3b^3c^3.a^6b^8c^9}\)

(do \(a\leq 1, b\leq 2\))

hay \(A\geq 4\sqrt[12]{(abc)^{12}}=4abc\)

Do đó ta có đpcm.

Dấu bằng xảy ra khi \((a,b,c)=(1,2,3)\)