Cho \(a,b,c>\dfrac{9}{4}.\)
Tìm \(MinP=\dfrac{a}{2\sqrt{b}-3}+\dfrac{b}{2\sqrt{c}-3}+\dfrac{c}{2\sqrt{a}-3}\)
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a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3}{\sqrt{x}-3}\)
\(C=\left(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}+4}\)
\(=\dfrac{-3}{2\sqrt{x}+4}\)
Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)
\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)
\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)
hay \(0\le x< \dfrac{25}{4}\)
\(\left(a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}\right)\left(1+3+5\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow3\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}\ge a+b+c\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{1}{a}+\dfrac{3^2}{b}+\dfrac{5^2}{c}}\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{\left(1+3+5\right)^2}{a+b+c}}=\dfrac{2}{3}\left(a+b+c\right)+\dfrac{27}{\sqrt{a+b+c}}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a+b+c\right)+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{1}{6}\left(a+b+c\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{27^2\left(a+b+c\right)}{2^3\left(a+b+c\right)}}+\dfrac{1}{6}.9=15\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;3;5\right)\)
b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)
\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)
c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)
d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
Áp dụng \(x^2+y^2+z^2\ge xy+yz+zx\) và \(x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^2\)
\(N\ge\dfrac{a^2b}{c}+\dfrac{b^2c}{a}+\dfrac{c^2a}{b}\ge\dfrac{1}{3}\left(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}\right)^2=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có:
\(ab+bc+ca\le\dfrac{1}{3}\left(a+b+c\right)^2=3\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+3}}\le\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự:
\(\dfrac{b}{\sqrt{b^2+3}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+3}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{c}{a+c}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)
\(P_{max}=\dfrac{3}{2}\) khi \(a=b=c=1\)
Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn
Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
\(M\ge\dfrac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{b}+\sqrt{c}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{c}+\sqrt{a}\right)^2}}{2}\)
\(M\ge\sqrt{a}+\sqrt{b}+\sqrt{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Lời giải:
Với điều kiện đã cho thì hiển nhiên mẫu dương.
Áp dụng BĐT Cauchy-Schwarz ta có:
\(M=\frac{a^2}{2a\sqrt{b}-3a}+\frac{b^2}{2b\sqrt{c}-3b}+\frac{c^2}{2c\sqrt{a}-3c}\)\(\geq \frac{(a+b+c)^2}{2(a\sqrt{b}+b\sqrt{c}+c\sqrt{a})-3(a+b+c)}\)
Áp dụng BĐT Bunhiacopxky kết hợp BĐT AM-GM:
\((a\sqrt{b}+b\sqrt{c}+c\sqrt{a})^2\leq (a+b+c)(ab+bc+ac)\)
\(\leq (a+b+c).\frac{(a+b+c)^2}{3}=\frac{(a+b+c)^3}{3}\)
\(\Rightarrow a\sqrt{b}+b\sqrt{c}+c\sqrt{a}\leq \sqrt{\frac{(a+b+c)^3}{3}}\)
\(\Rightarrow M\geq \frac{(a+b+c)^2}{2\sqrt{\frac{(a+b+c)^3}{3}}-3(a+b+c)}\)
Đặt \(\sqrt{\frac{a+b+c}{3}}=t(t>\frac{3}{2})\)\(\Rightarrow a+b+c=3t^2\)
Ta có:
\(P\geq\frac{9t^4}{6t^3-9t^2}=\frac{3t^2}{2t-3}\)
\(\Leftrightarrow P\geq \frac{\frac{3}{4}(2t-3)(2t+3)}{2t-3}+\frac{27}{4(2t-3)}\)
\(\Leftrightarrow P\geq \frac{3}{4}(2t+3)+\frac{27}{4(2t-3)}=\frac{3}{4}(2t-3)+\frac{27}{4(2t-3)}+\frac{9}{2}\)
Áp dụng BĐT AM-GM:
\(\frac{3}{4}(2t-3)+\frac{27}{4(2t-3)}\geq 2\sqrt{\frac{3}{4}.\frac{27}{4}}=\frac{9}{2}\)
\(\Rightarrow P\geq \frac{9}{2}+\frac{9}{2}=9\)
Vậy \(P_{\min}=9\)
Đặt \(\left\{{}\begin{matrix}\sqrt{a}=x\\\sqrt{b}=y\\\sqrt{c}=z\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{x^2}{2y-3}+\dfrac{y^2}{2z-3}+\dfrac{z^2}{2x-3}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)-9}\ge9\)
Vì \(\dfrac{t^2}{2t-9}-9=\dfrac{\left(t-9\right)^2}{2t-9}\ge0\) (với \(t=x+y+z\))