\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\supseteq\frac{a+b+c+d}{2}\left(a,b,c,d>0\right),\)
\(\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}< 5\left(a,b,c\supseteq0;a+b+c=1\right)\),
\(\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}< 6,5\)
\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\subseteq\sqrt{8}\left(a,b,c,d\supseteq0;a+b+c+d=1\right)\)
a) Áp dụng bdt cosi schwars ta có
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{a+b+b+c+c+d+d+a}\)
\(=\frac{a+b+c+d}{2}\)
bh mk can mn ho tro jup mk 2 cau cuoi nha
a) Áp dụng bđt Bunhiacopxki ta có :
\(\left[\left(\frac{a}{\sqrt{a+b}}^2\right)+\left(\frac{b}{\sqrt{b+c}}\right)^2+\left(\frac{c}{\sqrt{c+d}}\right)^2+\left(\frac{d}{\sqrt{d+a}}\right)\right]\)\(\times\)\(\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+d}\right)^2+\left(\sqrt{d+a}\right)^2\right]\)\(\ge\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\right)\times2\left(a+b+c+d\right)\ge\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge\frac{a+b+c+d}{2}\)( chia cả 2 vế cho \(2\left(a+b+c+d\right)\))
Dấu "=" xảy ra khi : a = b = c = d
Vậy ...
b) Áp dụng bđt Cauchy cho 2 số không âm ta có :
\(\sqrt{a+2}\le\frac{a+2+1}{2}\)
\(\sqrt{b+2}\le\frac{b+2+1}{2}\)
\(\sqrt{c+2}\le\frac{c+2+1}{2}\)
\(\Rightarrow\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\le\frac{a+b+c+2+1+2+1+2+1}{2}\)
Mà \(a+b+c=1\)
\(\Rightarrow\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}\le5\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}a+2=1\\b+2=1\\c+2=1\end{cases}}\) \(\Leftrightarrow a=b=c=-1\)
Mà \(a;b;c\ge0\)
Nên dấu "=" không xảy ra
Vậy \(\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}< 5\)
c) mạn phép sửa đề :) \(\left(a,b,c,d\ge0;a+b+c=1\right)\)
Áp dụng bđt Cauchy cho 2 số không âm ta có :
\(\sqrt{a+3}\le\frac{a+3+1}{2}\)
\(\sqrt{b+3}\le\frac{b+3+1}{2}\)
\(\sqrt{c+3}\le\frac{c+3+1}{2}\)
\(\Rightarrow\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\le\frac{a+b+c+12}{2}\)
Mà \(a+b+c=1\)
\(\Rightarrow VT\le6,5\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}c+3=1\\b+3=1\\c+3=1\end{cases}}\) \(\Leftrightarrow a=b=c=-2\)
Mà \(a+b+c=1\)
Nên dấu "=" không xảy ra
Vậy ...
d) Áp dụng bđt Bunhiacopxki ta có :
\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\right)^2\)\(\le\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+d}\right)^2+\left(\sqrt{d+a}\right)^2\right]\)\(\left(1+1+1+1\right)\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\right)^2\le\left(a+b+b+c+c+d+d+a\right)\times4\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\right)^2\le2\left(a+b+c+d\right)\times4\)
Mà \(a+b+c+d=1\)
\(\Leftrightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\right)^2\le8\)
\(\Leftrightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}\le\sqrt{8}\)
Dấu "=" xảy ra khi : \(a=b=c=d\)
Vậy ...
