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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{1}{x+1}=a,\dfrac{1}{y+1}=b,\dfrac{1}{z+1}=c\Rightarrow a,b,c>0;a+b+c=1.\)
\(x=\dfrac{1}{a}-1\)
Cần chứng minh: \(\sum\sqrt{\dfrac{1}{a}-1}\le\dfrac{3}{2}\sqrt{\left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right)}\)
Hay \(\sum\sqrt{\dfrac{1}{a}-\dfrac{1}{a+b+c}}\le\dfrac{3}{2}\sqrt{\prod\left(\dfrac{1}{a}-\dfrac{1}{a+b+c}\right)}\)
Hay là \(\sum\sqrt{\dfrac{b+c}{a\left(a+b+c\right)}}\le\dfrac{3}{2}\sqrt{\prod\dfrac{\left(b+c\right)}{a\left(a+b+c\right)}}\)
Tương đương: \(\sum\sqrt{\dfrac{b+c}{a}}\le\dfrac{3}{2}\sqrt{\prod\dfrac{\left(b+c\right)}{a}}\)
\(\left[\sum\left(b+c\right)\left\{a+2\left(b+c\right)\right\}\right]\left[\sum\dfrac{1}{a\left\{a+2\left(b+c\right)\right\}}\right]\ge\left[\sum\sqrt{\dfrac{b+c}{a}}\right]^2\)
Từ đây cần chứng minh:
\(\dfrac{9}{4}\prod\dfrac{\left(b+c\right)}{a}\ge\left[\sum\left(b+c\right)\left\{a+2\left(b+c\right)\right\}\right]\left[\sum\dfrac{1}{a\left\{a+2\left(b+c\right)\right\}}\right]\)
Còn lại bạn tự làm hoặc không để tối rảnh mình làm.
Do hoc24.vn không cho cập nhật câu trả lời nữa nên mình đăng tiếp:
Thực hiện thay thế \(\left(a,b,c\right)\rightarrow\left(s-a',s-b',s-c'\right)\) với $a',b',c'$ là độ dài ba cạnh của một tam giác.
Đặt $\left\{ \begin{array}{l}a' + b' + c' = 2s\\a'b' + b'c' + c'a' = {s^2} + 4Rr + {r^2}\\a'b'c' = 4sRr\end{array} \right.$
Bất đẳng thức quy về:
$${\dfrac { \left( 4\,R-24\,r \right) {s}^{4}+r \left( 72\,{R}^{2}+41\,Rr+8\,{r}^{2} \right) {s}^{2}+2\,{r}^{2} \left( 4\,R+r \right) ^{3}}{r{s}^{2} \left( 4\,{s}^{2}+r \left( 8\,R+r \right) \right) }}\geqslant 0$$
\( \Leftrightarrow \left( {4{\mkern 1mu} R - 24{\mkern 1mu} r} \right){s^4} + r\left( {72{\mkern 1mu} {R^2} + 41{\mkern 1mu} Rr + 8{\mkern 1mu} {r^2}} \right){s^2} + 2{\mkern 1mu} {r^2}{\left( {4{\mkern 1mu} R + r} \right)^3} \geqslant 0\)
Hay là \({s^2}\left( {R - 2{\mkern 1mu} r} \right)\left( {9{\mkern 1mu} {r^2} + 4{\mkern 1mu} {s^2}} \right) + r\left[ {10{\mkern 1mu} {s^2}\left( {4{\mkern 1mu} {R^2} + 4{\mkern 1mu} Rr + 3{\mkern 1mu} {r^2} - {s^2}} \right) + \left( {8{\mkern 1mu} Rr + 2{\mkern 1mu} {r^2} + 2{\mkern 1mu} {s^2}} \right)\left( {16{\mkern 1mu} {R^2} + 8{\mkern 1mu} Rr + {r^2} - 3{\mkern 1mu} {s^2}} \right)} \right] \geqslant 0\)
Đây là điều hiển nhiên.
Ngoài ra phương pháp SOS, SS cũng có thể sử dụng ở đây.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bổ đề:\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Ta có:\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)
Tương tự ta có:\(\dfrac{1}{2y+z+x}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)\)
\(\dfrac{1}{2z+x+y}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}\right)\)
Cộng vế với vế ta có:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{2y+z+x}+\dfrac{1}{2z+x+y}\le\dfrac{1}{16}\left[4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}.4.4=1\)
Dấu "=" xảy ra ⇔ \(x=y=z=\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\geq \frac{16}{3x+3y+2z}\)
\(\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\geq \frac{16}{3x+2y+3z}\)
\(\frac{1}{z+y}+\frac{1}{z+y}+\frac{1}{x+z}+\frac{1}{x+y}\geq \frac{16}{2x+3y+3z}\)
Cộng theo vế:
\(\Rightarrow 4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\geq 16\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)
\(\Rightarrow \frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\leq \frac{4.6}{16}=\frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{x+y-z}\Leftrightarrow \frac{x+y}{xy}=\frac{1}{z}+\frac{1}{x+y-z}=\frac{x+y}{z(x+y-z)}\)
\(\Leftrightarrow (x+y)(\frac{1}{xy}-\frac{1}{z(x+y-z)})=0\)
\(\Leftrightarrow (x+y).\frac{z(x+y-z)-xy}{xyz(x+y-z)}=0\)
\(\Leftrightarrow (x+y).\frac{(z-x)(y-z)}{xyz(x+y-z)}=0\)
\(\Leftrightarrow (x+y)(z-x)(y-z)=0\)
Xét các TH sau:
TH1: $x+y=0$. TH này loại do ĐKXĐ $x,y>0$
TH2: $z-x=0\Leftrightarrow z=x$
$\Leftrightarrow \frac{1}{y}=\frac{2020}{2021}$
\(M=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{y}}=2\sqrt{\frac{2020}{2021}}\)
TH3: $y-z=0$ tương tự TH2, ta có \(M=2\sqrt{\frac{2020}{2021}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta cần chứng minh:
\(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(1\right)\left(a,b>0\right)\)
\(\Leftrightarrow\dfrac{4}{a+b}\le\dfrac{a+b}{ab}\\ \Leftrightarrow4ab\le\left(a+b\right)^2\\ \Leftrightarrow\left(a-b\right)^2\ge0\left(luôn.đúng\right)\)
\(DBXR\Leftrightarrow a=b\)
Do các phép biến đổi tương đương nên (1) luôn đúng
Áp dụng (1), ta có:
\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}\left[\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Chứng minh tương tự, ta được:
\(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)
Cộng từng vế BĐT, ta được:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}.\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)Hay \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\left(đpcm\right)\)
\(DBXR\Leftrightarrow x=y=z=\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\)\(\Rightarrow\left\{{}\begin{matrix}a+b+c=1\\a;b;c>0\end{matrix}\right.\)
Và \(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}+\dfrac{bc}{\sqrt{b^2+c^2+2a^2}}+\dfrac{ca}{\sqrt{c^2+a^2+2b^2}}\le\dfrac{1}{2}\)
Ta có:\(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}=\dfrac{2ab}{\sqrt{\left(1+1+2\right)\left(a^2+b^2+2c^2\right)}}\)
\(\le\dfrac{2ab}{a+b+2c}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{bc+ac}{a+b}\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)
Dấu "=" khi \(a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=\dfrac{1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{x^2+yz}+\dfrac{y}{y^2+zx}+\dfrac{z}{z^2+xy}\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}+\dfrac{1}{\sqrt{xy}}\right)\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\).
Đẳng thức xảy ra khi x = y = z = 1.
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)\geq (1+1+1)^2\)
\(\Leftrightarrow \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)\geq 9\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}> \frac{4}{x+y+z}\)
Vậy BĐT đã cho được cm. Dấu bằng không xảy ra .