![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2 là tìm giá trị lớn nhất ạ!
ta có A>=0. xét 100=xy+z+xz\(\ge3\sqrt[3]{xy\cdot yz\cdot zx}\)
\(\Rightarrow100\ge3\sqrt[3]{A^2}\Rightarrow\left(\frac{100}{3}\right)^3\ge A^2\Rightarrow A< \frac{100}{3}\sqrt{\frac{100}{3}}\)
dấu đẳng thức xảy ra khi xy=yz=zx
Bài 1 nhìn vô đoán ngay a=3,b=2 -> S=13!
AM-GM:\(\frac{5}{9}\left(a^2+9\right)\ge\frac{10}{3}a;\text{ }\frac{4}{9}\left(a^2+\frac{9}{4}b^2\right)\ge\frac{4}{3}ab\)
\(\rightarrow a^2+b^2+5\ge\frac{10}{3}a+\frac{4}{3}ab\ge\frac{10}{3}\cdot3+\frac{4}{3}\cdot6=18\)
\(\Rightarrow S=a^2+b^2\ge13\) (đúng)
Đẳng thức xảy ra khi a=3, b=2.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=2\)
=> \(\dfrac{1}{a+1}=1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}=\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge2\sqrt{\dfrac{bc}{\left(b+1\right)\left(c+1\right)}}\)( AM-GM)
Tương tự ta có \(\dfrac{1}{b+1}\ge2\sqrt{\dfrac{ac}{\left(a+1\right)\left(c+1\right)}}\); \(\dfrac{1}{c+1}\ge2\sqrt{\dfrac{ab}{\left(a+1\right)\left(b+1\right)}}\)
Nhân vế với vế các bđt trên
=> \(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge8\sqrt{\dfrac{a^2b^2c^2}{\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2}}=8\cdot\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)
=> \(1\le8abc\)<=> \(abc\le\dfrac{1}{8}\)
Đẳng thức xảy ra <=> a=b=c=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1+a}=1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{bc}{\left(1+b\right)\left(1+c\right)}}\)
Tương tự:
\(\dfrac{1}{1+b}\ge2\sqrt{\dfrac{ac}{\left(1+a\right)\left(1+c\right)}}\) ; \(\dfrac{1}{1+c}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+c\right)}}\)
Nhân vế với vế:
\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Rightarrow abc\le\dfrac{1}{8}\)
\(N_{max}=\dfrac{1}{8}\) khi \(a=b=c=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
=> BDT cần CMR <=> \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\ge\frac{a^2}{a^2+b^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+a^2}\)
Ta có \(\frac{a^3}{a^2+b^2}=a-\frac{ab^2}{a^2+b^2}\ge a-\frac{ab^2}{2ab}=a-\frac{b}{2}\)
=>VT\(\ge\frac{a+b+c}{2}\) (Hơi tắt nên tự hiểu)
Ta đi Cm \(\frac{a+b+c}{2}\ge\frac{a^2}{a^2+b^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+a^2}\)
<=> \(\frac{a+b+c}{2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{c^2+b^2}+\frac{a^2}{a^2+c^2}\ge3\)(*)
\(\frac{a+b+c}{2}\ge\frac{3}{2}\)
\(\frac{b^2}{a^2+b^2}+\frac{c^2}{c^2+b^2}+\frac{a^2}{c^2+a^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2b^2+b^2c^2+c^2a^2\right)}\ge\frac{3}{2}\)
=>VT (*) \(\ge3\). Từ đó ta có dpcm
Kiêm đâu lắm bài bdt hay. Gửi link
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Tìm min:
Áp dụng BĐT AM-GM:
$x^2+y^2+z^2\geq \frac{(x+y+z)^2}{3}=\frac{6^2}{3}=12$
Vậy $A_{\min}=12$. Giá trị này đạt tại $x=y=z=2$
--------------
Tìm max:
$A=x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=36-2(xy+yz+xz)$
Vì $x,y,z\geq 0\Rightarrow xy+yz+xz\geq 0$
$\Rightarrow A=36-2(xy+yz+xz)\leq 36$
Vậy $A_{\max}=36$. Giá trị này đạt tại $(x,y,z)=(0,0,6)$ và hoán vị.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P^2=\left(a-b\right)^2\left(b-c\right)^2\left(a-c\right)^2\)
Không mất tính tổng quát, giả sử \(c=min\left\{a;b;c\right\}\) \(\Rightarrow\left\{{}\begin{matrix}\left(b-c\right)^2\le b^2\\\left(a-c\right)^2\le a^2\end{matrix}\right.\)
\(\Rightarrow P^2\le\left(a-b\right)^2a^2b^2=\dfrac{1}{4}\left(a^2-2ab+b^2\right).\left(2ab\right).\left(2ab\right)\le\dfrac{1}{108}\left(a^2-2ab+b^2+2ab+2ab\right)^3\)
\(\Rightarrow P^2\le\dfrac{1}{108}\left(a+b\right)^6\le\dfrac{1}{108}\left(a+b+c\right)^6=\dfrac{27}{4}\)
\(\Rightarrow P\le\dfrac{3\sqrt{3}}{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3-\sqrt{3}}{2};\dfrac{3+\sqrt{3}}{2};0\right)\) và các hoán vị
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Theo bất đẳng thức Cô - si, ta có: \(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\le\frac{a+b+a+c}{2}+\frac{b+c}{2}=1\)
\(\Rightarrow\sqrt{a}\left(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\right)\le\sqrt{a}\)hay \(\sqrt{a^2+abc}+\sqrt{abc}\le\sqrt{a}\)
Tương tự ta có: \(\sqrt{b^2+abc}+\sqrt{abc}\le\sqrt{b}\);\(\sqrt{c^2+abc}+\sqrt{abc}\le\sqrt{c}\)
Mà \(abc\le\left(\frac{a+b+c}{3}\right)^3=\frac{1}{27}\Rightarrow\sqrt{abc}\le\frac{1}{3\sqrt{3}}\)
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\le3\left(a+b+c\right)=3\)\(\Leftrightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\le\sqrt{3}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)
\(P=\frac{ab\frac{c}{2}\frac{c}{2}}{\frac{c}{4}}\le\frac{4\left(a+b+c\right)^4}{3.256}=\frac{27}{4}\)
Áp dụng BĐT cô-si cho 2 số không âm ta có:
\(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\)
\(\Leftrightarrow2\ge\sqrt[3]{abc}\)
\(\Leftrightarrow abc\le8\)
chắc GTLN là 8 heee ko rành