Bài 1:

Áp dụng BĐT Bunhiacopxky ta có:

$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$

$\Leftrightarrow 3(a^2+b^2+c^2)\geq 1$

$\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

Bài 2: 

Áp dụng BĐT Bunhiacopxky:

$(a^2+4b^2+9c^2)(1+\frac{1}{4}+\frac{1}{9})\geq (a+b+c)^2$

$\Leftrightarrow 2015.\frac{49}{36}\geq (a+b+c)^2$

$\Leftrightarrow \frac{98735}{36}\geq (a+b+c)^2$

$\Rightarrow a+b+c\leq \frac{7\sqrt{2015}}{6}$ chứ không phải $\frac{\sqrt{14}}{6}$ :''>>

Bài 3:

Áp dụng BĐT Bunhiacopxky:

$2=(a^2+b^2)(1+1)\geq (a+b)^2\Rightarrow a+b\leq \sqrt{2}$

$(a\sqrt{1+a}+b\sqrt{1+b})^2\leq (a^2+b^2)(1+a+1+b)$

$=2+a+b\leq 2+\sqrt{2}$

$\Rightarrow a\sqrt{1+a}+b\sqrt{1+b}\leq \sqrt{2+\sqrt{2}}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=\frac{1}{\sqrt{2}}$

Bài 3:

\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)

\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)

-Tham khảo:

-Tham khảo:

\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+c^3=\dfrac{3}{abc}\)

\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{3}{abc}=0\)

\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{1}{c}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c^2}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)

\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{1}{ab}-\dfrac{1}{bc}-\dfrac{1}{ca}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b=c\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\)

Đề bài thiếu, cần thêm dữ liệu "a;b;c phân biệt"

Khi đó \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)

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

d) => 2a^2 + 2b^2 + 2c^2 = 2ab+ 2bc + 2ca

    => 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0

( a^2 - 2ab+b^2 ) + ( a^2 - 2ac + c^2) + ( b^2 - 2bc - c^2) = 0

(a-b)^2 + (a-c)^2 + (b-c)^2 = 0

=> | ( a-b)^2 = 0 => a=b     
     |  ( a-c)^2 = 0 => a=c
     |  ( b-c)^2 = 0 => b=c

=>>> a=b=c

b) => 2(a-b)^2 - (a-b)^2  = 0

   2 ( a^2- 2ab + b^2) - a^2+ 2ab - b^2 = 0

  2a^2 - 4ab+ 2b^2 - a^2 + 2ab - b^2 = 0

 a^2 -2ab + b^2 =0 

( a-b)^2 = 0 => a=b

Cái này bạn nên xem lại đề có đúng ko nha~~ Mk ko lm ra số đối đc Sorry

b) => (a-b)^2 - 2(a-b)^2 = 0

   a^2 - 2ab+b^2 - 2( a^2 -2ab+ b^2) = 0

 a^2 - 2ab +b^2 - 2a^2 + 4ab - 2b^2 = 0

-a^2 - 2ab -b^2 = 0

- (a+b)^2 = 0

=> (a+b)^2 = 0 => a= -b ( đpcm)

\(\Leftrightarrow a\left(a+2\right)\left(c+2\right)+b\left(a+2\right)\left(c+2\right)+c\left(b+2\right)\left(c+2\right)\le\left(a+2\right)\left(b+2\right)\left(c+2\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+ab^2+bc^2+ca^2\le8+abc\)

\(\Leftrightarrow ab^2+bc^2+ca^2\le2+abc\)

Không mất tính tổng quát, giả sử \(b=mid\left\{a;b;c\right\}\)

\(\Rightarrow\left(a-b\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow ab+bc\ge b^2+ac\)

\(\Leftrightarrow ab^2+ca^2\le a^2b+abc\)

\(\Rightarrow ab^2+bc^2+ca^2\le bc^2+a^2b+abc=b\left(a^2+c^2\right)+abc=b\left(2-b^2\right)+abc\)

\(=2+abc-\left(b-1\right)^2\left(b+2\right)\le2+abc\) (đpcm)