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\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
CMTT: \(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ac}{b+c}+\dfrac{ac}{b+a}\right)\)
\(\Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{ab}{c+a}+\dfrac{ab}{c+b}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ac}{b+c}+\dfrac{ac}{b+c}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left[\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right]=\dfrac{1}{2}\left(a+b+c\right)=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(P=\frac{\sqrt{ab}}{(a+c)+(b+c)}+\frac{\sqrt{bc}}{(b+a)+(c+a)}+\frac{\sqrt{ca}}{(c+b)+(a+b)}\)
\(\leq \underbrace{\frac{\sqrt{ab}}{2\sqrt{(a+c)(b+c)}}+\frac{\sqrt{bc}}{2\sqrt{(b+a)(c+a)}}+\frac{\sqrt{ca}}{2\sqrt{(c+b)(a+b)}}}_{M}(*)\)
Xét:
\(M=\frac{1}{2}\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}(1)\)
Theo BĐT Bunhiacopxky và AM-GM:
\((\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)})^2\leq (ab+bc+ac)(a+b+b+c+c+a)\)
\(=2(ab+bc+ac)(a+b+c)=2[(a+b)(b+c)(c+a)+abc]\)
\(\leq 2[(a+b)(b+c)(c+a)+\frac{(a+b)(b+c)(c+a)}{8}]=\frac{9}{4}(a+b)(b+c)(c+a)\)
\(\Rightarrow \sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}\leq \frac{3}{2}\sqrt{(a+b)(b+c)(c+a)}(2)\)
Từ \((1);(2)\Rightarrow M\leq \frac{1}{2}.\frac{3}{2}=\frac{3}{4}(**)\)
Từ \((*); (**)\Rightarrow P\leq M\leq \frac{3}{4}\)
Vậy \(P_{\max}=\frac{3}{4}\Leftrightarrow a=b=c\)
Bunhiacopxki:
\(\left(b+a+a\right)\left(b+c+\dfrac{c^2}{a}\right)\ge\left(b+\sqrt{ca}+c\right)^2\)
\(\Rightarrow\dfrac{2a^2+ab}{\left(b+\sqrt{ca}+c\right)^2}\ge\dfrac{2a^2+ab}{\left(2a+b\right)\left(b+c+\dfrac{c^2}{a}\right)}=\dfrac{a^2}{c^2+ab+bc}\)
Tương tự:
\(\dfrac{2b^2+bc}{\left(c+\sqrt{ca}+a\right)^2}\ge\dfrac{b^2}{a^2+ab+bc}\)
\(\dfrac{2c^2+ca}{\left(a+\sqrt{bc}+b\right)^2}\ge\dfrac{c^2}{b^2+ac+bc}\)
\(\Rightarrow P\ge\dfrac{a^2}{c^2+ab+ac}+\dfrac{b^2}{a^2+ab+bc}+\dfrac{c^2}{b^2+ac+bc}\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)
Dấu "=" xảy ra khi \(a=b=c\)
Lợi dụng Cauchy-Schwarz' inequality ta có:
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{ab+ac+bc+c^2}}\)
\(=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự ta cũng có:
\(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ca}{\sqrt{ca+2b}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{bc+ca}{a+b}+\dfrac{ab+ca}{b+c}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{b+c}\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\cdot2=1\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)
Ta có P=\(\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}+\dfrac{bc}{\sqrt{bc+\left(a+b+c\right)a}}+\dfrac{ac}{\sqrt{ac+\left(a+b+c\right)b}}\)
=\(\dfrac{ab}{\sqrt{ab+ac+bc+c^2}}+\dfrac{bc}{\sqrt{bc+ac+ab+a^2}}+\dfrac{ac}{\sqrt{ac+ab+bc+b^2}}\)
=\(\dfrac{ab}{\sqrt{a\left(b+c\right)+c\left(b+c\right)}}+\dfrac{bc}{\sqrt{b\left(a+c\right)+a\left(a+c\right)}}+\dfrac{ac}{\sqrt{c\left(a+b\right)+b\left(a+b\right)}}\)
=\(\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}+\dfrac{bc}{\sqrt{\left(b+a\right)\left(c+a\right)}}+\dfrac{ac}{\sqrt{\left(a+b\right)\left(c+b\right)}}\)
áp dụng bđt Cói ta có:
\(\sqrt{\left(a+c\right)\left(b+c\right)}\)\(\le\)\(\dfrac{2+c}{2}=1+\dfrac{c}{2}\)
\(\sqrt{\left(b+á\right)\left(c+a\right)}\)
Ta có: $\sqrt[]{ab+2c}=\sqrt[]{ab+(a+b+c)c}=\sqrt[]{ab+ac+bc+c^2}=\sqrt[]{(c+a)(c+b)}$ (do $a+b+c=2$)
Nên $\dfrac{ab}{\sqrt[]{ab+2c}}=\dfrac{ab}{\sqrt[]{(c+a).(c+b)}}=ab.\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}}$
Áp dụng bất đẳng thức Cauchy cho $\dfrac{1}{a+c};\dfrac{1}{b+c}>0$ có:
$\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}} \leq \dfrac{1}{2}.(\dfrac{1}{a+c}+\dfrac{1}{b+c})$
Nên $\dfrac{ab}{\sqrt[]{ab+2c}} \leq \dfrac{1}{2}.ab.(\dfrac{1}{a+c}+\dfrac{1}{b+c})= \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})$
Tương tự ta có: $\dfrac{bc}{\sqrt[]{bc+2a}} \leq \dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})$
$\dfrac{ca}{\sqrt[]{ca+2b}} \leq \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})$
Nên $Q \leq \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})+\dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})+ \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}.(\dfrac{b(a+c)}{a+c}+\dfrac{a(b+c)}{b+c}+\dfrac{c(a+b)}{a+b}=\dfrac{1}{2}.(a+b+c)=1$ (do $a+b+c=2$)
Dấu $=$ xảy ra khi $a=b=c=\dfrac{2}{3}$
Theo BĐT \(AM-GM\) ta có :
\(\dfrac{a}{\sqrt{2b^2+2c^2-a^2}}=\dfrac{\sqrt{3}a^2}{\sqrt{3a^2\left(2b^2+2c^2-a^2\right)}}\ge\dfrac{\sqrt{3}a^2}{\dfrac{2a^2+2b^2+2c^2}{2}}=\dfrac{\sqrt{3}a^2}{a^2+b^2+c^2}\)
Tương tự ta có :
\(\dfrac{b}{\sqrt{2c^2+2a^2-b^2}}\ge\dfrac{\sqrt{3}b^2}{a^2+b^2+c^2}\)
\(\dfrac{c}{\sqrt{2a^2+2b^2-c^2}}\ge\dfrac{\sqrt{3}c^2}{a^2+b^2+c^2}\)
Cộng từng vế BĐT :
\(\Rightarrow VT\ge\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=\sqrt{3}\)
\("="\Leftrightarrow a=b=c\)
\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)
\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)
Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)
Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
\(a+b+c=\sqrt{6063}\Leftrightarrow\dfrac{a}{\sqrt{2021}}+\dfrac{b}{\sqrt{2021}}+\dfrac{c}{\sqrt{2021}}=\sqrt{3}\)
Đặt \(\left(\dfrac{a}{\sqrt{2021}};\dfrac{b}{\sqrt{2021}};\dfrac{c}{\sqrt{2021}}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{3}\)
\(P=\dfrac{2x}{\sqrt{2x^2+1}}+\dfrac{2y}{\sqrt{2y^2+1}}+\dfrac{2z}{\sqrt{2z^2+1}}\)
Ta có đánh giá:
\(\dfrac{x}{\sqrt{2x^2+1}}\le\dfrac{3\sqrt{15}x+2\sqrt{5}}{25}\)
Thật vậy, BĐT tương đương:
\(\left(\sqrt{3}x-1\right)^2\left(9x^2+10\sqrt{3}x+2\right)\ge0\) (luôn đúng)
Tương tự và cộng lại:
\(P\le\dfrac{6\sqrt{15}\left(x+y+z\right)+12\sqrt{5}}{25}=\dfrac{6\sqrt{5}}{5}\)
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Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
\(\dfrac{\sqrt{ab}}{a+c+b+c}\le\dfrac{\sqrt{ab}}{2\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{4}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
Tương tự và cộng lại:
\(A\le\dfrac{1}{4}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)