![](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)
Lời giải:
Áp dụng BĐT Cauchy:
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{a^2+b^2}=\frac{a^2+b^2}{a^2b^2}+\frac{4}{a^2+b^2}\geq 2\sqrt{\frac{a^2+b^2}{a^2b^2}.\frac{4}{a^2+b^2}}=\frac{4}{ab}=\frac{32(a^2+b^2)}{8ab(a^2+b^2)}(1)\)
Tiếp tục áp dụng BĐT Cauchy ngược dấu:
\(8ab(a^2+b^2)=4.(2ab).(a^2+b^2)\leq (2ab+a^2+b^2)^2=(a+b)^4(2)\)
Từ \((1);(2)\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{4}{a^2+b^2}\geq \frac{32(a^2+b^2)}{8ab(a^2+b^2)}\geq \frac{32(a^2+b^2)}{(a+b)^4}\) (đpcm)
Dấu "=" xảy ra khi $a=b$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ab+\frac{1}{a^2}+\frac{1}{b^2}\ge ab+\frac{2}{ab}=ab+\frac{1}{16ab}+\frac{31}{16ab}\ge2\sqrt{\frac{ab}{16ab}}+\frac{31}{\frac{16\left(a+b\right)^2}{4}}\ge\frac{1}{2}+\frac{31}{4}=\frac{33}{4}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cauchy -Schwarz dạng cộng mẫu thôi:
\(\text{VT}=\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}+\frac{4^2}{d}\geq \frac{(1+1+2+4)^2}{a+b+c+d}=\frac{64}{a+b+c+d}=\text{VP}\)
Dấu bằng xảy ra khi \(a=b=\frac{c}{2}=\frac{d}{4}>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải
Ta có: \(\left(a+b+\frac{1}{4}\right)^2=\frac{1}{16}\left(4a+4b-1\right)^2+\left(a+b\right)\ge a+b\)
Tương tự: \(\left(b+c+\frac{1}{4}\right)^2\ge b+c;\left(c+a+\frac{1}{4}\right)^2\ge c+a\)
Như vậy: \(L.H.S\left(VT\right)\ge\left(a+b\right)+\left(b+c\right)+\left(c+a\right)=\left(\frac{1}{\frac{1}{a}}+\frac{1}{\frac{1}{b}}\right)+\left(\frac{1}{\frac{1}{b}}+\frac{1}{\frac{1}{c}}\right)+\left(\frac{1}{\frac{1}{c}}+\frac{1}{\frac{1}{a}}\right)\)
\(\ge4\left(\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\right)=R.H.S\left(VP\right)\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{8}\). Ta có đpcm.
khác cách tth xíu
Ta có:
\(VP=\Sigma_{cyc}\frac{4}{\frac{1}{a}+\frac{1}{b}}\le\Sigma_{cyc}\frac{4}{\frac{4}{a+b}}=2\left(a+b+c\right)\)
Gio ta di chung minh
\(VT\ge2\left(a+b+c\right)\)
Ta lai co:
\(VT=\Sigma_{cyc}\left(a+b+\frac{1}{4}\right)^2\ge\frac{\left[2\left(a+b+c\right)+\frac{3}{4}\right]^2}{3}\)
Chung minh
\(\frac{\left[2\left(a+b+c\right)+\frac{3}{4}\right]^2}{3}\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\left[2\left(a+b+c\right)-\frac{3}{4}\right]^2\ge0\) (đúng)
Dau '=' xay ra khi \(a=b=c=\frac{1}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng bđt \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)(bđt svacxo) ta có :
VT= \(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge\frac{\left(1+1+2+4\right)^2}{a+b+c+d}\)= \(\frac{64}{a+b+c+d}\)=VP (đpcm)
dấu = xảy ra <=>a=b=1; c=2 ; d=4
Dễ dàng CM BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b},\forall a,b>0\)
Áp dụng liên tục ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge\frac{4}{a+b}+\frac{4}{c}+\frac{16}{d}\ge4.\frac{4}{a+b+c}+\frac{16}{d}\ge16.\frac{4}{a+b+c+d}=\frac{64}{a+b+c+d}\)
dấu = xảy ra <=> a+b=c, a+b+c=d, a=b
ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)
\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)
b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)
\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)
Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)
Cộng vế với vế ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(0\le a,b,c\le1\)
nên\(\left\{{}\begin{matrix}\left(a^2-1\right)\left(b-1\right)\ge0\\\left(b^2-1\right)\left(c-1\right)\ge0\\\left(c^2-1\right)\left(a-1\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2b-b-a^2+1\ge0\\b^2c-c-b^2+1\ge0\\c^2a-a-c^2+1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2b\ge a^2+b-1\\b^2c\ge b^2+c-1\\c^2a\ge c^2+a-1\end{matrix}\right.\)
Ta cũng có:
\(2\left(a^3+b^3+c^3\right)\le a^2+b+b^2+c+c^2+a\)
Do đó \(T=2\left(a^3+b^3+c^3\right)-\left(a^2b+b^2c+c^2a\right)\)
\(\le a^2+b+b^2+c+c^2+a\)\(-\left(a^2+b-1+b^2+c-1+c^2+a-1\right)\)
\(=3\)
Vậy GTLN của T=3, đạt được chẳng hạn khi \(a=1;b=0;c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{abc}\left(a+b+c\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Từ đó suy ra đpcm