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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(P=\frac{a}{a+2bc}+\frac{b}{b+2ca}+\frac{c}{c+2ab}\)
\(\Leftrightarrow P=\frac{a^2}{a^2+2bca}+\frac{b^2}{b^2+2cab}+\frac{c^2}{c^2+2abc}\)
Áp dụng BĐT Cauchy-schwarz ta có: ( link c/m Cauchy-schwarz: Xem câu hỏi )
\(P\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+6abc}=\frac{9}{a^2+b^2+c^2+6abc}\)( \(a+b+c=3\))
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Ta có: \(a+b+c=3\)
Áp dụng BĐT AM-GM ta có:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\Leftrightarrow3\ge3\sqrt[3]{abc}\)
\(\Leftrightarrow1\ge\sqrt[3]{abc}\)
\(\Leftrightarrow1\ge abc\)
\(\Leftrightarrow a^2b^2c^2\ge a^3b^3c^3\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Áp dụng BĐT AM-GM ta có:
\(ab+bc+ca\ge3.\sqrt[3]{a^2b^2c^2}\ge3.\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
\(\Rightarrow P\ge\frac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}=\frac{9}{9}=1\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\sqrt{a^2+2ab+b^2+b^2}+\sqrt{b^2+2bc+c^2+c^2}+\sqrt{c^2+2ca+a^2+a^2}\)
\(M=\sqrt{\left(a+b\right)^2+b^2}+\sqrt{\left(b^{ }+c\right)^2+c^2}+\sqrt{\left(c+a\right)^2+a^2}\)
\(M\ge\sqrt{\left(a+b+b+c+c+a\right)^2+\left(a+b+c\right)^2}\ge\sqrt{\left[2\left(a+b+c\right)\right]^2+3^2}\ge\sqrt{6^2+3^2}\ge3\sqrt{5}\)
\(dấu\)\("="xảy\) \(ra\) \(\Leftrightarrow a=b=c=1\)
Cách khác:
Áp dụng BĐT Bunhiacopxky:
$5(a^2+2ab+2b^2)=[(a+b)^2+b^2](2^2+1^2)\geq [2(a+b)+b]^2$
$\Rightarrow \sqrt{5(a^2+2ab+b^2)}\geq 2a+3b$
Tương tự với các căn thức còn lại và cộng theo vế:
$M\sqrt{5}\geq 5(a+b+c)$
$\Leftrightarrow M\geq \sqrt{5}(a+b+c)=3\sqrt{5}$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta sẽ chứng minh :
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) với x, y > 0
Thật vậy : \(x+y+z\ge3\sqrt[3]{xyz}\)( bđt Cô - si )
Và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\) ( bđt Cô - si )
\(\Rightarrow x+y+z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) ( Dấu " = " \(\Leftrightarrow x=y=z\) )
Ta có :
\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
( Dấu " = " xay ra khi a=b)
Tương tự ta cũng có :
\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\) ( Dấu " = " xảy ra khi b=c)
\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\) ( Dấu " = " xay ra khi c = a )
\(VT=\sum_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
Dấu " = " xay ra khi \(a=b=c=\frac{2}{3}\)
Chúc bạn học tốt !!
\(\frac{1}{\sqrt{4a^2+2ab+b^2+a^2+b^2}}\le\frac{1}{\sqrt{4a^2+2ab+b^2+2ab}}=\frac{1}{\sqrt{\left(2a+b\right)^2}}=\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow VT\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}+\frac{2}{b}+\frac{1}{c}+\frac{2}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(a^3+1+1+b^3+1+1+c^3+1+1\ge3\left(a+b+c\right)\)
\(\Rightarrow3\left(a+b+c\right)\le a^3+b^3+c^3+6\le9\)
\(\Rightarrow a+b+c\le3\)
\(\Rightarrow ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\le3\)
Quay lại bài toán ta có:
\(\left(\frac{ab}{\sqrt{c+3}}+\frac{bc}{\sqrt{a+3}}+\frac{ca}{\sqrt{b+3}}\right)^2\le\left(ab+bc+ca\right)\left(\frac{ab}{c+3}+\frac{bc}{a+3}+\frac{ca}{b+3}\right)\)
\(\le3.\left(\frac{ab}{c+3}+\frac{bc}{a+3}+\frac{ca}{b+3}\right)\)
\(\le3.\left(\frac{ab}{c+a+c+b}+\frac{bc}{a+b+a+c}+\frac{ca}{b+a+b+c}\right)\)
\(\le\frac{3}{4}.\left(\frac{ab}{c+a}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{c+a}+\frac{ca}{a+b}+\frac{ca}{b+c}\right)\)
\(=\frac{3}{4}.\left(\frac{ca}{a+b}+\frac{bc}{a+b}+\frac{bc}{c+a}+\frac{ab}{c+a}+\frac{ca}{b+c}+\frac{ab}{b+c}\right)\)
\(=\frac{3}{4}.\left(a+b+c\right)\le\frac{9}{4}\)
\(\Rightarrow\frac{ab}{\sqrt{c+3}}+\frac{bc}{\sqrt{a+3}}+\frac{ca}{\sqrt{b+3}}\le\frac{3}{2}\)
\(\Rightarrow\frac{2ab}{\sqrt{c+3}}+\frac{2bc}{\sqrt{a+3}}+\frac{2ca}{\sqrt{b+3}}\le3\)
PS: Được chưa 2 cô nương Hoàng Lê Bảo Ngọc, Witch Rose.
Số t khổ quá mà. Thấy có bài giải mừng húm tưởng khỏi cần giải nữa thì vô đọc bài của bác Thắng Nguyễn thấy mệt mệt. Bác lo mà úp mặt vô tường đi :(
Cái này xấu lắm đấy nhé :v, chủ thớt muốn thì post thôi @@
*)Note:\(Σ\) là tổng đối xứng viết tắt cho gọn
\(\text{∏}\) tích đối xứng viết tắt luôn :v \(\text{∏}a=abc;Σa=a+b+c\)
\(BDT\Leftrightarrow\frac{ab}{\sqrt{c+3}}+\frac{bc}{\sqrt{a+3}}+\frac{ca}{\sqrt{b+3}}\le\frac{3}{2}\)
Theo Cauchy-Schwarz và đặt \(a+b+c=3u;ab+bc+ca=3v^2;abc=w^3\)
\(\left(Σ\frac{ab}{\sqrt{c+3}}\right)^2\leΣab\cdotΣ\frac{ab}{c+3}\le\frac{9}{4}\)
\(\Leftrightarrow\frac{3v^2Σab\left(a+3\right)\left(b+c\right)}{\text{∏}\left(a+3\right)}\le\frac{9}{4}\)
\(\Leftrightarrow4v^2Σ\left(a^2b^2+3a^2b+3a^2c+9ab\right)\le3\left(abc+27+Σ\left(3ab+9a\right)\right)\)
\(\Leftrightarrow4v^2\left(9v^4-6uw^3+27uv^2-9w^3+27v^2\right)\le3\left(w^3+9v^2+27+27u\right)\)
\(\Leftrightarrow w^3\left(1+12v^2+8uv^2\right)+27u+27+9v^2\ge12v^6+36uv^4+36v^4\)
A[ dụng BDT Schur có:\(w^3\ge4uv^2-3u^3\)
Nên cần cm \(\left(4uv^2-3u^3\right)\left(1+12v^2+8uv^2\right)+27u+27+9v^2\ge12v^6+36uv^4+36v^4\)
\(\Leftrightarrow32u^2v^4+12uv^4+4uv^2+9v^2+27u+27\ge12v^6+36v^4+3u^3+24u^2v^2+36u^3v^2\)
Đúng theo BĐT P-M và BĐT AM-GM
P.s: Đọc đến đây thì cho hỏi cái đề đâu ra thế, thật sự lm ko muốn dùng cách này đâu @@ hại não, hại mắt
![](https://rs.olm.vn/images/avt/0.png?1311)
với mọi x,y,z >0 ta có: \(x+y+z\ge3\sqrt[3]{xyz};\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)
\(\Rightarrow\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
đẳng thức xảy ra khi x=y=z
ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
đẳng thức xảy ra khi a=b
tương tự: \(\frac{1}{\sqrt{5b^2+2ab+2b^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
đẳng thức xảy ra khi b=c
\(\frac{1}{\sqrt{5c^2+2bc+2c^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
đẳng thức xảy ra khi c=a
Vậy \(\frac{1}{\sqrt{5a^2+2ca+2a^2}}+\frac{1}{\sqrt{5b^2+2bc+2c^2}}+\frac{1}{\sqrt{5c^2+2ac+2a^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
đẳng thức xảy ra khi a=b=c=\(\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta viết lại bất đẳng thức cần chứng minh thành\(\sqrt{\frac{2\left(a+3\right)}{a+bc}}+\sqrt{\frac{2\left(b+3\right)}{b+ca}}+\sqrt{\frac{2\left(c+3\right)}{c+ab}}\ge6\)
Theo giả thiết, ta có a + b + c = 3 nên\(\sqrt{\frac{2\left(a+3\right)}{a+bc}}=\sqrt{\frac{2\left(a+a+b+c\right)}{a+bc}}=\sqrt{2\left(\frac{a+b}{a+bc}+\frac{a+c}{a+bc}\right)}\)\(\ge\sqrt{\frac{a+b}{a+bc}}+\sqrt{\frac{a+c}{a+bc}}\)(Áp dụng bất đẳng thức \(\sqrt{2\left(x+y\right)}\ge\sqrt{x}+\sqrt{y}\))
Hoàn toàn tương tự, ta được: \(\sqrt{\frac{2\left(b+3\right)}{b+ca}}\ge\sqrt{\frac{b+a}{b+ca}}+\sqrt{\frac{b+c}{b+ca}}\); \(\sqrt{\frac{2\left(c+3\right)}{c+ab}}\ge\sqrt{\frac{c+a}{c+ab}}+\sqrt{\frac{c+b}{c+ab}}\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(\sqrt{\frac{2\left(a+3\right)}{a+bc}}+\sqrt{\frac{2\left(b+3\right)}{b+ca}}+\sqrt{\frac{2\left(c+3\right)}{c+ab}}\)\(\ge\sqrt{\frac{a+b}{a+bc}}+\sqrt{\frac{a+c}{a+bc}}+\sqrt{\frac{b+a}{b+ca}}+\sqrt{\frac{b+c}{b+ca}}+\sqrt{\frac{c+a}{c+ab}}+\sqrt{\frac{c+b}{c+ab}}\)
Áp dụng bất đẳng thức Bunyakovsky dạng phân thức, ta được: \(\sqrt{\frac{a+b}{a+bc}}+\sqrt{\frac{a+b}{b+ca}}\ge\frac{4\sqrt{a+b}}{\sqrt{a+bc}+\sqrt{b+ca}}\ge\frac{2\sqrt{2}\sqrt{a+b}}{\sqrt{a+bc+b+ca}}=\frac{2\sqrt{2}}{\sqrt{c+1}}\)(*)
Tương tự ta có: \(\sqrt{\frac{b+c}{b+ca}}+\sqrt{\frac{b+c}{c+ab}}\ge\frac{2\sqrt{2}}{\sqrt{a+1}}\)(**) ; \(\sqrt{\frac{c+a}{c+ab}}+\sqrt{\frac{c+a}{a+bc}}\ge\frac{2\sqrt{2}}{\sqrt{b+1}}\)(***)
Cộng theo vế ba bất đẳng thức (*), (**) và (***) suy ra \(\sqrt{\frac{a+b}{a+bc}}+\sqrt{\frac{a+c}{a+bc}}+\sqrt{\frac{b+a}{b+ca}}+\sqrt{\frac{b+c}{b+ca}}+\sqrt{\frac{c+a}{c+ab}}+\sqrt{\frac{c+b}{c+ab}}\)\(\ge\frac{2\sqrt{2}}{\sqrt{c+1}}+\frac{2\sqrt{2}}{\sqrt{a+1}}+\frac{2\sqrt{2}}{\sqrt{b+1}}\)
Do đó ta có: \(\sqrt{\frac{2\left(a+3\right)}{a+bc}}+\sqrt{\frac{2\left(b+3\right)}{b+ca}}+\sqrt{\frac{2\left(c+3\right)}{c+ab}}\ge\frac{2\sqrt{2}}{\sqrt{c+1}}+\frac{2\sqrt{2}}{\sqrt{a+1}}+\frac{2\sqrt{2}}{\sqrt{b+1}}\)
Phép chứng minh sẽ hoàn tất nếu ta chỉ ra được \(\frac{2\sqrt{2}}{\sqrt{c+1}}+\frac{2\sqrt{2}}{\sqrt{a+1}}+\frac{2\sqrt{2}}{\sqrt{b+1}}\ge6\)hay \(\frac{1}{\sqrt{c+1}}+\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}\ge\frac{3}{\sqrt{2}}\)
Thật vậy, áp dụng bất đẳng thức Cauchy – Schwarz ta được \(\frac{1}{\sqrt{c+1}}+\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}\ge\frac{9}{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}}\ge\frac{9}{\sqrt{3\left(a+b+c+3\right)}}=\frac{3}{\sqrt{2}}\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi a = b = c = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta sẽ chứng minh: \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)với x,y > 0.
Thật vậy: \(x+y+z\ge3\sqrt[3]{xyz}\)(bđt Cô -si)
và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\)(bđt Cô -si)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)(Dấu "="\(\Leftrightarrow x=y=z\))
Ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
(Dấu "=" xảy ra khi a = b)
Tương tự ta có:\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)(Dấu "=" xảy ra khi b=c)
\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)(Dấu "=" xảy ra khi c=a)
\(VT=\text{Σ}_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
(Dấu "=" xảy ra khi \(a=b=c=\frac{3}{2}\))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}\le\dfrac{1}{\sqrt{\left(2a+b\right)^2}}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự ta có: \(\dfrac{1}{\sqrt{5b^2+2bc+2c^2}}\le\dfrac{1}{9}\left(\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\dfrac{1}{\sqrt{5c^2+2ac+a^2}}\le\dfrac{1}{9}\left(\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế với vế:
\(\dfrac{1}{\sqrt{5a^2+2ab+b^2}}+\dfrac{1}{\sqrt{5b^2+2bc+c^2}}+\dfrac{1}{\sqrt{5c^2+2ac+a^2}}\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\le\dfrac{2}{3}\)
Dấu "=" khi \(a=b=c=\dfrac{3}{2}\)
Áp dụng bđt: 2xy \(\le\)(x + y)2/2
khi đó, ta có: \(\sqrt{\frac{a+b}{2ab}}\ge\sqrt{\frac{a+b}{\frac{\left(a+b\right)^2}{2}}}=\sqrt{\frac{2}{a+b}}=\frac{1}{\sqrt{\frac{a+b}{2}}}\ge\frac{1}{\frac{\frac{a+b}{2}+1}{2}}=\frac{4}{a+b+2}\)
CMTT: \(\sqrt{\frac{b+c}{2bc}}\ge\frac{4}{b+c+2}\)
\(\sqrt{\frac{c+a}{2ca}}\ge\frac{4}{c+a+2}\)
=>Đặt A = \(\sqrt{\frac{a+b}{2ab}}+\sqrt{\frac{b+c}{2bc}}+\sqrt{\frac{a+c}{2ac}}\ge\frac{4}{a+b+2}+\frac{4}{b+c+2}+\frac{4}{a+c+2}\)
Áp dụng bđt svacso : \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\)
ta có:
\(A\ge\frac{\left(2+2+2\right)^2}{a+b+2+b+c+2+a+c+2}=\frac{36}{2\left(a+b+c\right)+6}=\frac{36}{12}=3\)
=> Đpcm