• Ta có :

 \(\hept{\begin{cases}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{cases}}\) \(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\Rightarrow a^2+b^2+c^2\ge ab+bc+ac\)

• Theo bất đẳng thức tam giác : 

​\(\hept{\begin{cases}a+b>c\\b+c>a\\a+c>b\end{cases}}\)\(\Rightarrow\hept{\begin{cases}c\left(a+b\right)>c^2\\a\left(b+c\right)>a^2\\b\left(a+c\right)>b^2\end{cases}}\) \(\Rightarrow\hept{\begin{cases}c^2< bc+ac\\a^2< ab+ac\\b^2< ab+bc\end{cases}}\) \(\Rightarrow a^2+b^2+c^2< 2\left(ab+bc+ac\right)\)

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

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

Tương tự và cộng lại:

\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)

Ta có: a + b + c = 2 nên \(2c+ab=c\left(a+b+c\right)+ab=ac+bc+c^2+ab\)

\(=\left(ca+c^2\right)+\left(bc+ab\right)=c\left(a+c\right)+b\left(a+c\right)\)\(=\left(b+c\right)\left(a+c\right)\)

Áp dụng BĐT Cô - si cho 2 số không âm:

\(\frac{1}{b+c}+\frac{1}{a+c}\ge2\sqrt{\frac{1}{\left(b+c\right)\left(a+c\right)}}\)(Vì a,b,c thực dương)

\(\Rightarrow\sqrt{\frac{1}{\left(b+c\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+c}\right)\)

\(\Rightarrow\frac{1}{\sqrt{2c+ab}}\le\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+c}\right)\)(cmt)

\(\Rightarrow\frac{ab}{\sqrt{ab+2c}}\le\frac{1}{2}\left(\frac{ab}{b+c}+\frac{ab}{a+c}\right)\)(nhân 2 vế cho ab thực dương)    (1)

(Dấu "="\(\Leftrightarrow\frac{1}{b+c}=\frac{1}{c+a}\Leftrightarrow b+c=c+a\Leftrightarrow a=b\))

Tương tự ta có: \(\frac{bc}{\sqrt{bc+2a}}\le\frac{1}{2}\left(\frac{bc}{b+a}+\frac{bc}{a+c}\right)\)(Dấu "="\(\Leftrightarrow b=c\))  (2)

\(\frac{ca}{\sqrt{ca+2b}}\le\frac{1}{2}\left(\frac{ca}{c+b}+\frac{ca}{b+a}\right)\)(Dấu "="\(\Leftrightarrow a=c\))  (3)

Cộng các BĐT (1) , (2) , (3), ta được:

\(P\le\frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}+\frac{bc}{b+a}+\frac{cb}{c+a}+\frac{ac}{b+a}+\frac{ac}{c+b}\right)\)

\(\Rightarrow P\le\frac{1}{2}\left(\frac{b\left(c+a\right)}{c+a}+\frac{a\left(c+b\right)}{c+b}+\frac{c\left(b+a\right)}{b+a}\right)\)

\(\le\frac{1}{2}\left(a+b+c\right)=1\)

Vậy \(P=\frac{ab}{\sqrt{ab+2c}}\)\(+\frac{bc}{\sqrt{bc+2a}}\)\(+\frac{ca}{\sqrt{ca+2b}}\le1\)

(Dấu "="\(\Leftrightarrow a=b=c=\frac{2}{3}\))

Ta có:

\(\frac{ab}{\sqrt{ab+2c}}=\frac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{ab}{c+a}+\frac{ab}{c+b}\)

Tương tự:

\(\frac{bc}{\sqrt{bc+2a}}\le\frac{bc}{a+b}+\frac{bc}{a+c}\)

\(\frac{ca}{\sqrt{ca+2b}}\le\frac{ca}{b+c}+\frac{ca}{b+a}\)

Khi đó:

\(P\le\frac{ab}{a+c}+\frac{ab}{c+b}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{b+c}+\frac{ca}{b+a}\)

\(=\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}+\frac{c\left(a+b\right)}{b+a}\)

\(=a+b+c=2\)

Dấu "=" xảy ra tại \(a=b=c=\frac{2}{3}\)

Ta có : \(ab+bc+ac\le a^2+b^2+c^2\Leftrightarrow2\left(ab+bc+ac\right)\le2\left(a^2+b^2+c^2\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Vì BĐT cuối luôn đúng nên ta có : \(a^2+b^2+c^2\ge ab+bc+ac\)

Theo Bất đẳng thức tam giác ta có : 

\(a< b+c\Rightarrow a.a< a\left(b+c\right)\Leftrightarrow a^2< ab+ac\) (1)

\(b< a+c\Rightarrow b.b< b\left(a+c\right)\Leftrightarrow b^2< ab+bc\)(2)

\(c< a+b\Rightarrow c.c< c\left(a+b\right)\Leftrightarrow c^2< ac+bc\)(3)

Cộng (1) , (2) , (3) theo vế ta được : \(a^2+b^2+c^2< 2\left(ab+bc+ac\right)\)

Từ đó suy ra đpcm

Nếu em lên lớp 7 thì em sẽ giúp

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

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

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

+ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow ab+b+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)

\(\Rightarrow ab+bc+ca< \frac{1}{2}\)

\(VT=\sqrt{\frac{ab+2c^2}{a^2+ab+b^2}}+\sqrt{\frac{bc+2a^2}{b^2+bc+c^2}}+\sqrt{\frac{ca+2b^2}{c^2+ca+a^2}}\)

\(=\frac{ab+2c^2}{\sqrt{\left(a^2+ab+b^2\right)\left(ab+2c^2\right)}}+\frac{bc+2a^2}{\sqrt{\left(b^2+bc+c^2\right)\left(bc+2a^2\right)}}+\frac{ca+2b^2}{\sqrt{\left(c^2+ca+a^2\right)\left(ca+2b^2\right)}}\)

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

\(\ge\frac{ab+2c^2}{a^2+b^2+c^2}+\frac{bc+2a^2}{a^2+b^2+c^2}+\frac{ca+2b^2}{a^2+b^2+c^2}=ab+bc+ca+2\left(a^2+b^2+c^2\right)\)

\(=2+ab+bc+ca=VP\) (Do a² + b² + c² = 1) => ĐPCM.

Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{\sqrt{3}}.\)

chăc là .............................. điền đi sẽ biếc a you ok ?

\(\sum\dfrac{a}{b^2+bc+c^2}\ge\dfrac{\left(a+b+c\right)^2}{ab^2+abc+ac^2+bc^2+abc+ba^2+ca^2+abc+cb^2}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}=\dfrac{a+b+c}{ab+bc+ac}\)

Đúng rầu đấy

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

\(=\left(a-bc\right)\sqrt{\frac{1}{\left(a+b\right)^2\left(c+a\right)^2}}\le\frac{\frac{a-bc}{\left(a+b\right)^2}+\frac{a-bc}{\left(c+a\right)^2}}{2}=\frac{a-bc}{2\left(a+b\right)^2}+\frac{a-bc}{2\left(c+a\right)^2}\)

Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)

=> \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{a-bc+b-ca}{2\left(a+b\right)^2}+\frac{b-ca+c-ab}{2\left(b+c\right)^2}+\frac{a-bc+c-ab}{2\left(c+a\right)^2}\)

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

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)