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Áp dụng BĐT Bunhia:
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{\left(1+1+1\right)\left(4a+1+4b+1+4c+1\right)}\)
\(\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{3.\left(4\left(a+b+c\right)+3\right)}=\sqrt{21}< \sqrt{25}=5\)
Vậy \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5\)
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
\(a.\) Áp dụng BĐT Cô - Si cho các số không âm , ta có :
\(\sqrt{1}.\sqrt{a+1}\le\dfrac{a+1+1}{2}=\dfrac{a+2}{2}\)
\(\sqrt{1}.\sqrt{b+1}\le\dfrac{b+1+1}{2}=\dfrac{b+2}{2}\)
\(\sqrt{1}.\sqrt{c+1}\le\dfrac{c+1+1}{2}=\dfrac{c+2}{2}\)
\(\Rightarrow\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le\dfrac{a+b+c+6}{2}=\dfrac{7}{2}=3,5\)
Dấu \("="\) xảy ra khi : \(\left\{{}\begin{matrix}a+1=1\\b+1=1\\c+1=1\end{matrix}\right.\)\(\Leftrightarrow a=b=c=0\)\(\Rightarrow a+b+c\ne1\left(trái-với-giả-thiết\right)\)
\(\Rightarrow\) Dấu \("="\) không xảy ra .
\(\Rightarrow\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}< 3,5\)
\(b.\) Áp dụng BĐT Bunhiacopxki , ta có :
\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}\right)^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+a+c\right)=3.2=6\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}\le\sqrt{6}\)
Dấu " = " xảy ra khi : \(a+b=b+c=a+c\Rightarrow a=b=c=\dfrac{1}{3}\)
Câu a : Dùng BĐT Bu-nhi-a-cốp-xki ta có :
\(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le\sqrt{3\left(a+b+c+3\right)}=\sqrt{12}=3,46< 3,5\)
Câu b tương tự :
\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6\left(a+b+c\right)}=\sqrt{6}\)
ta có:\(a,b,c\ge0;a+b+c=4\)
\(\Rightarrow a+b\le4\)\(mà\)\(a,b\ge0\)\(\Rightarrow0\le a+b\le4\left(1\right)\)
\(\Rightarrow\sqrt{a+b}\le2\)
\(\Rightarrow2-\sqrt{a+b}\ge0\)\(\left(2\right)\)
Từ (1) và(2)\(\Rightarrow\sqrt{a+b}\left(2-\sqrt{a+b}\right)\ge0\)
\(\Rightarrow2\sqrt{a+b}\ge a+b\)
CMTT:\(2\sqrt{b+c}\ge b+c;2\sqrt{c+a}\ge c+a\)
\(\Rightarrow2\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\ge2\left(a+b+c\right)\)
Mà a+b+c=4\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge4\)
Dấu "="xảy ra khi \(\left(a;b;c\right)=\left(4;0;0\right);\left(0;4;0\right);\left(0;0;4\right)\)