\(\sqrt{a^2+ab+2b^2}=\sqrt{\left(\frac{3}{4}a+\frac{5}{4}b\right)^2+\frac{7}{16}\left(a-b\right)^2}\ge\sqrt{\left(\frac{3}{4}a+\frac{5}{4}b\right)^2}=\frac{3a+5b}{4}\)

Tương tự \(\sqrt{b^2+2c^2+bc}\ge\frac{3b+5c}{4};\sqrt{c^2+2a^2+ca}\ge\frac{3c+5a}{4}\)

\(\Rightarrow\sqrt{a^2+ab+2b^2}+\sqrt{b^2+2c^2+bc}+\sqrt{c^2+2a^2+ca}\ge\frac{3a+5b+3b+5c+3c+5a}{4}\)

\(=2\left(a+b+c\right)\left(đpcm\right)\)

\(a^2+2b^2+ab=\frac{7}{16}\left(a-b\right)^2+\frac{9}{16}\left(a+\frac{5}{3}b\right)^2\)

\(\Leftrightarrow\sqrt{a^2+2b^2+ab}=\sqrt{\frac{7}{16}\left(a-b\right)^2+\frac{9}{16}\left(a+\frac{5}{3}b\right)^2}\ge\sqrt{\frac{9}{16}\left(a+\frac{5}{3}b\right)^2}=\frac{3}{4}\left(a+\frac{5}{3}b\right)\)

Tương tự \(\sqrt{b^2+2c^2+bc}\ge\frac{3}{4}\left(b+\frac{5}{3}c\right),\sqrt{c^2+2a^2+ac}\ge\frac{3}{4}\left(c+\frac{5}{3}a\right)\)

Cộng lại vế theo vế ta được: 

\(\sqrt{a^2+2b^2+ab}+\sqrt{b^2+2c^2+bc}+\sqrt{c^2+2a^2+ca}\ge\frac{3}{4}\left(a+\frac{5}{3}b+b+\frac{5}{3}c+c+\frac{5}{3}a\right)\)

\(=2\left(a+b+c\right)\).

Dấu \(=\)khi \(a=b=c\ge0\).

Còn cách khác nè :

Đặt \(P=\sqrt{a^2+2b^2+ab}+\sqrt{b^2+2c^2+bc}+\sqrt{c^2+2a^2+ac}\)

Ta chứng minh \(P\ge2\left(a+b+c\right)\)

\(2P=\sqrt{\left(1+1+2\right)\left(a^2+2b^2+ab\right)}+\sqrt{\left(1+1+2\right)\left(b^2+2c^2+bc\right)}+\sqrt{\left(1+1+2\right)\left(c^2+2a^2+ac\right)}\)

Áp dụng bđt bunyakovsky ta được:

\(2P\ge a+2b+\sqrt{ab}+b+2c+\sqrt{bc}+c+2a+\sqrt{ac}\)

      \(=3\left(a+b+c\right)+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\ge4\left(a+b+c\right)\left(AM-GM\right)\)

Suy ra \(P\ge2\left(a+b+c\right)\left(đpcm\right)\)

mọi người làm cách tối cổ quá , cách tổng quát luôn này 

Ta cần cm \(\sqrt{xa^2+yab+zb^2}\ge ma+nb\)

Nếu \(x=z=>m=n=\frac{\sqrt{x+y+z}}{2}\)

Nếu \(x\ne z=>\hept{\begin{cases}m+n=\sqrt{x+y+z}\\m-n=\frac{x-z}{\sqrt{x+y+z}}\end{cases}}\)

Áp dụng : \(\sqrt{a^2+ab+2b^2}\ge ma+nb\)

Với \(x=1;y=1;z=2\)

Vì \(x\ne z\)\(=>\hept{\begin{cases}m+n=\sqrt{x+y+z}\\m-n=\frac{x-z}{\sqrt{x+y+z}}\end{cases}}\)

\(< =>\hept{\begin{cases}m+n=\sqrt{4}\\m-n=-\frac{1}{\sqrt{4}}\end{cases}}\)

\(< =>\hept{\begin{cases}m+n=\sqrt{4}\\2m=\sqrt{4}-\frac{1}{\sqrt{4}}\end{cases}}\)

\(< =>\hept{\begin{cases}m+n=2\\m=1-\frac{1}{4}=\frac{3}{4}\end{cases}}\)

\(< =>\hept{\begin{cases}m=\frac{3}{4}\\n=\frac{5}{4}\end{cases}}\)

Nên ta cần chứng minh \(\sqrt{a^2+ab+2b^2}\ge\frac{3}{4}a+\frac{5}{4}b\)

đến đây thì bình phương 2 vế rồi chuyển vế là được bđt đúng nhé 

Lời giải:

$a^2+2b^2+ab=\frac{a^2}{2}+\frac{3b^2}{2}+\frac{(a+b)^2}{2}$
Áp dụng BĐT Bunhiacopxky:

$[\frac{a^2}{2}+\frac{3b^2}{2}+\frac{(a+b)^2}{2}](2+6+8)\geq (a+3b+2a+2b)^2$

$\Rightarrow \sqrt{a^2+2b^2+ab}\geq \frac{3a+5b}{4}$

Hoàn toàn tương tự với các căn còn lại suy ra:
$\text{VT}\geq \frac{3a+5b}{4}+\frac{3b+5c}{4}+\frac{3c+5a}{4}=2(a+b+c)$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

Bạn xem lại đề xem có nhầm không?

a/ Nếu (a + b) < 0 thì bất  đẳng thức đúng

Với (a + b) \(\ge0\)thì ta có

\(2a^2+ab+2b^2\ge\frac{5}{4}\left(a^2+2ab+b^2\right)\)

\(\Leftrightarrow3a^2-6ab+3b^2\ge0\)

\(\Leftrightarrow3\left(a-b\right)^2\ge0\)(đúng)

b/ Áp dụng BĐT BCS : 

\(1=\left(1.\sqrt{a}+1.\sqrt{b}+1.\sqrt{c}\right)^2\le3\left(a+b+c\right)\Rightarrow a+b+c\ge\frac{1}{3}\)

Áp dụng câu a/ :

\(\sqrt{2a^2+ab+2b^2}\ge\frac{\sqrt{5}}{2}\left(a+b\right)\)

\(\sqrt{2b^2+bc+2c^2}\ge\frac{\sqrt{5}}{2}\left(b+c\right)\)

\(\sqrt{2c^2+ac+2a^2}\ge\frac{\sqrt{5}}{2}\left(a+c\right)\)

\(\Rightarrow P\ge\frac{\sqrt{5}}{2}.2\left(a+b+c\right)\ge\frac{\sqrt{5}}{3}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{9}\)

Vậy min P = \(\frac{\sqrt{5}}{3}\) khi a=b=c=1/9

\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)

\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)

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 \(a=b=c=\dfrac{1}{\sqrt{3}}\)

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)

Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.

Lời giải vắn tắt:

\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)

( thay \(a^2+b^2+c^2=1\))

\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)

Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)

Cộng vế với vế:

\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)