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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 \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
\(a+b+c=\sqrt{6063}\Leftrightarrow\dfrac{a}{\sqrt{2021}}+\dfrac{b}{\sqrt{2021}}+\dfrac{c}{\sqrt{2021}}=\sqrt{3}\)
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:
\(P\le\dfrac{6\sqrt{15}\left(x+y+z\right)+12\sqrt{5}}{25}=\dfrac{6\sqrt{5}}{5}\)
Áp dụng bđt Cauchy Shwarz và bđt phụ \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow M^2=\left(\sqrt{\dfrac{a}{b+c+2a}}+\sqrt{\dfrac{b}{c+a+2b}}+\sqrt{\dfrac{c}{a+b+2c}}\right)^2\)
\(\le\left(1+1+1\right)\left(\dfrac{a}{b+c+2a}+\dfrac{b}{c+a+2b}+\dfrac{c}{a+b+2c}\right)\)
\(\le\dfrac{3}{4}\left(\dfrac{a}{b+a}+\dfrac{a}{c+a}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{9}{4}\)
➤ \(M\le\dfrac{3}{2}\)
Dấu "=" xảy ra ⇔ a = b = c
\(M=\sqrt{\dfrac{a}{b+c+2a}}+\sqrt{\dfrac{b}{c+a+2b}}+\sqrt{\dfrac{c}{a+b+2c}}\)
\(\le\dfrac{1}{4}+\dfrac{a}{b+c+2a}+\dfrac{1}{4}+\dfrac{b}{c+a+2b}+\dfrac{1}{4}+\dfrac{c}{a+b+2c}\)
\(\le\dfrac{3}{4}+\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
\(=\dfrac{3}{4}+\dfrac{1}{4}.\left(1+1+1\right)=\dfrac{3}{2}\)
Ta có \(ab+bc+ca=3abc\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) thì ta có \(x,y,z>0;x+y+z=3\) và
\(\sqrt{\dfrac{a}{3b^2c^2+abc}}=\sqrt{\dfrac{\dfrac{1}{x}}{3.\dfrac{1}{y^2z^2}+\dfrac{1}{xyz}}}=\sqrt{\dfrac{\dfrac{1}{x}}{\dfrac{3x+yz}{xy^2z^2}}}=\sqrt{\dfrac{y^2z^2}{3x+yz}}\) \(=\dfrac{yz}{\sqrt{3x+yz}}\) \(=\dfrac{yz}{\sqrt{x\left(x+y+z\right)+yz}}\) \(=\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)
Do đó \(T=\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)
Lại có \(\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{yz}{2\left(x+y\right)}+\dfrac{yz}{2\left(x+z\right)}\)
Lập 2 BĐT tương tự rồi cộng theo vế, ta được \(T\le\dfrac{yz}{2\left(x+y\right)}+\dfrac{yz}{2\left(x+z\right)}+\dfrac{zx}{2\left(y+z\right)}+\dfrac{zx}{2\left(y+x\right)}\) \(+\dfrac{xy}{2\left(z+x\right)}+\dfrac{xy}{2\left(z+y\right)}\)
\(T\le\dfrac{yz+zx}{2\left(x+y\right)}+\dfrac{xy+zx}{2\left(y+z\right)}+\dfrac{xy+yz}{2\left(z+x\right)}\)
\(T\le\dfrac{x+y+z}{2}\) (do \(x+y+z=3\))
\(T\le\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\) \(\Leftrightarrow a=b=c=1\)
Vậy \(maxT=\dfrac{3}{2}\), xảy ra khi \(a=b=c=1\)
(Mình muốn gửi lời cảm ơn tới bạn Nguyễn Đức Trí vì ý tưởng của bài này chính là bài mình vừa hỏi lúc nãy trên diễn đàn. Cảm ơn bạn Trí rất nhiều vì đã giúp mình có được lời giải này.)
Bạn Lê Song Phương xem lại dùm nhé, thanks!
\(...\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\)
\(...\Rightarrow T\le2.3=6\)
\(\Rightarrow GTLN\left(T\right)=6\left(tạia=b=c=1\right)\)
ad bunhiacopxki ta có
A^2 \(\le3\left(\dfrac{a}{b+c+2a}+\dfrac{b}{c+a+2c}+\dfrac{c}{a+b+2c}\right)\)
Đặt B=\(\dfrac{a}{b+c+2a}+\dfrac{b}{c+a+2b}+\dfrac{c}{a+b+2c}\)
\(\Leftrightarrow\)B-3 =-\(\left(a+b+c\right)\) \(\left(\dfrac{1}{b+c+2a}+\dfrac{1}{c+a+2b}+\dfrac{1}{a+b+2a}\right)\)
dễ CM \(\dfrac{1}{a+b+2c}+\dfrac{1}{b+c+2a}+\dfrac{1}{c+a+2b}\)\(\ge\dfrac{9}{4\left(a+c+b\right)}\)
\(\Rightarrow\)B-3\(\le\)\(\dfrac{-9}{4}\)\(\Rightarrow\)B\(\le\dfrac{3}{4}\)
\(\Rightarrow A^2\le\dfrac{9}{4}\) mà A>0
\(\Rightarrow\)A\(\le\dfrac{3}{2}\)Dấu = xra khi a=b=c
Lời giải:
Đặt biểu thức đã cho là $A$
Ta có:
\(A=\sqrt{\frac{a}{b+c+2a}}+\sqrt{\frac{b}{a+c+2b}}+\sqrt{\frac{c}{a+b+2c}}\)
\(A=\sqrt{\frac{a}{(a+b)+(a+c)}}+\sqrt{\frac{b}{(b+c)+(b+a)}}+\sqrt{\frac{c}{(c+a)+(c+b)}}\)
Áp dụng BĐT AM-GM:
\(A\leq\sqrt{\frac{a}{2\sqrt{(a+b)(a+c)}}}+\sqrt{\frac{b}{2\sqrt{(b+c)(b+a)}}}+\sqrt{\frac{c}{2\sqrt{(c+a)(c+b)}}}\)
\(\Leftrightarrow A\leq \sqrt[4]{\frac{a^2}{4(a+b)(a+c)}}+\sqrt[4]{\frac{b^2}{4(b+c)(b+a)}}+\sqrt[4]{\frac{c^2}{4(c+a)(c+b)}}(*)\)
Tiếp tục áp dụng AM-GM:
\(\sqrt[4]{\frac{a^2}{4(a+b)(a+c)}}\leq \frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{1}{2}+\frac{1}{2}\right)\)
\(\sqrt[4]{\frac{b^2}{4(b+c)(b+a)}}\leq \frac{1}{4}\left(\frac{b}{b+c}+\frac{b}{a+b}+\frac{1}{2}+\frac{1}{2}\right)\)
\(\sqrt[4]{\frac{c^2}{4(c+a)(c+b)}}\leq \frac{1}{4}\left(\frac{c}{c+a}+\frac{c}{c+b}+\frac{1}{2}+\frac{1}{2}\right)\)
Cộng theo vế kết hợp với $(*)$
\(\Rightarrow A\leq \frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+6.\frac{1}{2}\right)\)
\(\Leftrightarrow A\leq \frac{1}{4}.6=\frac{3}{2}\)
Vậy \(A_{\max}=\frac{3}{2}\Leftrightarrow a=b=c\)
\(a=b=c\rightarrow P=\frac{3}{2}\). Ta se c/m do la gtln của P. Thật vậy:
\(\frac{1}{2}P=\sqrt{\frac{1}{4}.\frac{a}{b+c+2a}}+...\)
\(\le\frac{1}{2}\left(\frac{1}{4}+\frac{a}{b+c+2a}+\frac{1}{4}+\frac{b}{c+a+2b}+\frac{1}{4}+\frac{c}{a+b+2c}\right)\)
\(=\frac{1}{2}\left(\frac{3}{4}+\frac{a}{\left(b+a\right)+\left(c+a\right)}+\frac{b}{\left(c+b\right)+\left(b+a\right)}+\frac{c}{\left(c+a\right)+\left(c+b\right)}\right)\)
\(\le\frac{1}{2}\left[\frac{3}{4}+\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)\right]=\frac{3}{4}\)
Do đó \(P\le\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c