Cho a≥25 b ≥36 c≥ 49 và a+ b + c= 125 . Tìm GTNN của H =\(\sqrt{a}+\sqrt{b}+\sqrt{c}\)
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Đặt \(\left(2\sqrt{a}-5;2\sqrt{b}-5;2\sqrt{c}-5\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x;y;z>0\\a=\left(\dfrac{x+5}{2}\right)^2\\b=\left(\dfrac{y+5}{2}\right)^2\\c=\left(\dfrac{z+5}{2}\right)^2\end{matrix}\right.\)
\(Q=\dfrac{\left(x+5\right)^2}{4y}+\dfrac{\left(y+5\right)^2}{4z}+\dfrac{\left(z+5\right)^2}{4x}\ge\dfrac{\left(x+y+z+15\right)^2}{4\left(x+y+z\right)}\)
\(Q\ge\dfrac{\left(x+y+z\right)^2+30\left(x+y+z\right)+225}{4\left(x+y+z\right)}\)
\(Q\ge\dfrac{x+y+z}{4}+\dfrac{225}{4\left(x+y+z\right)}+\dfrac{15}{2}\ge2\sqrt{\dfrac{225\left(x+y+z\right)}{16\left(x+y+z\right)}}+\dfrac{15}{2}=15\)
Dấu "=" xảy ra khi \(a=b=c=25\)
Áp dụng bđt hoán vị cho hai bộ số đơn điệu ngược chiều \(\left(a,b,c\right);\left(2\sqrt{a}-5,2\sqrt{b}-5,2\sqrt{c}-5\right)\): \(Q\ge\dfrac{a}{2\sqrt{a}-5}+\dfrac{b}{2\sqrt{b}-5}+\dfrac{c}{2\sqrt{c}-5}\).
Mặt khác ta có \(\dfrac{a}{2\sqrt{a}-5}-5=\dfrac{\left(\sqrt{a}-5\right)^2}{2\sqrt{a}-5}\ge0\).
Do đó \(Q\ge5+5+5=15\).
Dấu bằng xảy ra khi a = b = c = 25.
Ta có:
\(S=\dfrac{a^2}{a\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{b^2}{b\left(\sqrt{c}+\sqrt{a}\right)}+\dfrac{c^2}{c\left(\sqrt{a}+\sqrt{b}\right)}\ge\dfrac{\left(a+b+c\right)^2}{a\left(\sqrt{b}+\sqrt{c}\right)+b\left(\sqrt{c}+\sqrt{a}\right)+c\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(b+c\right)+\sqrt{b}\left(c+a\right)+\sqrt{c}\left(a+b\right)}\)
Mặt khác:
\(\sqrt{a}\left(b+c\right)=\dfrac{1}{\sqrt{2}}\sqrt{2a.\left(b+c\right)\left(b+c\right)}\le\dfrac{1}{\sqrt{2}}\sqrt{\left(\dfrac{2a+2b+2c}{3}\right)^3}=\dfrac{2\sqrt{3}}{9}\)
\(\Rightarrow S\ge\dfrac{1}{3.\dfrac{2\sqrt{3}}{9}}=\dfrac{\sqrt{3}}{2}\)
\(M\ge\dfrac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{b}+\sqrt{c}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{c}+\sqrt{a}\right)^2}}{2}\)
\(M\ge\sqrt{a}+\sqrt{b}+\sqrt{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta đặt:
\(\left\{{}\begin{matrix}x=a-1\\y=b-2\\z=c-3\end{matrix}\right.\)
\(\Rightarrow x+y+z=3\) và \(x,y,z\ge0\) (*)
Biểu thứ P trở thành:
\(P=\sqrt{x}+\sqrt{y}+\sqrt{z}\)
Từ (*) dễ thấy:
\(\left\{{}\begin{matrix}0\le x\le3\\0\le y\le3\\0\le z\le3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\le x\le\sqrt{3x}\\0\le y\le\sqrt{3y}\\0\le z\le\sqrt{3z}\end{matrix}\right.\)
Do đó:
\(P\ge\dfrac{x+y+z}{\sqrt{3}}=\sqrt{3}\)
Dầu "=" xảy ra khi \(\left(a;b;c\right)=\left(3;0;0\right)=\left(0;3;0\right)=\left(0;0;3\right)\)
Áp dụng BĐT cosi, ta có
\(\sqrt{3a+1}=\dfrac{1}{2}\sqrt{4\left(3a+1\right)}\le\dfrac{1}{2}.\dfrac{4+3a+1}{2}=\dfrac{3a+5}{4}\)
CMTT, ta có \(\sqrt{3b+1}\le\dfrac{3b+5}{4};\sqrt{3c+1}\le\dfrac{3c+5}{4}\)
Từ đó suy ra \(K\le\dfrac{3\left(a+b+c\right)+15}{4}=6\)
Dấu "=" xảy ra khi a=b=c=1
Vậy...
ta có BĐT \(\sqrt{3a+1}\ge\dfrac{a\left(\sqrt{10}-1\right)}{3}+1\)
\(\Leftrightarrow a\left(3-a\right)\ge0đúng\forall a\)
CMRTT, ta có
\(\sqrt{3b+1}\ge\dfrac{b\left(\sqrt{10}-1\right)}{3}+1\)
\(\sqrt{3c+1}\ge\dfrac{c\left(\sqrt{10}-1\right)}{3}+1\)
Do đó \(K\ge\dfrac{\left(a+b+c\right)\left(\sqrt{10}-1\right)}{3}+3=\sqrt{10}+2\)
Dấu "=" xảy ra khi a=3, b=c=0
Vậy...