\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)

Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

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Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)

Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)

- Nếu \(abc\ge0\Rightarrow a^2+b^2+c^2+abc\ge0\) dấu "=" xảy ra khi và chỉ khi \(a=b=c=0\)

- Nếu \(abc< 0\Rightarrow\)  trong 3 số a; b; c có ít nhất 1 số âm

Không mất tính tổng quát, giả sử \(c< 0\Rightarrow ab>0\)

Mà \(\left\{{}\begin{matrix}-2\le c< 0\\ab>0\end{matrix}\right.\Leftrightarrow abc\ge-2ab\)

\(\Rightarrow a^2+b^2+c^2+abc\ge a^2+b^2-2ab+c^2=\left(a-b\right)^2+c^2>0\) (không thỏa mãn)

Vậy \(a=b=c=0\)

a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b

Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)

Dấu = xảy ra <=> a=b=1

b) Áp dụng BĐT bunhiacopxki có:

\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)

c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)

Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) vào S ta được:

\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)

Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)

\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)

\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)

\(a^2+b^2\ge2ab\Rightarrow ab\le\dfrac{a^2+b^2}{2}\)

\(\Rightarrow4=a^2+b^2-ab\ge a^2+b^2-\dfrac{a^2+b^2}{2}=\dfrac{a^2+b^2}{2}\)

\(\Rightarrow a^2+b^2\le8\)

\(a^2+b^2\ge-2ab\Rightarrow-ab\le\dfrac{a^2+b^2}{2}\)

\(\Rightarrow4=a^2+b^2-ab\le a^2+b^2+\dfrac{a^2+b^2}{2}=\dfrac{3\left(a^2+b^2\right)}{2}\)

\(\Rightarrow\dfrac{8}{3}\le a^2+b^2\)

\(\Rightarrow\dfrac{8}{3}\le a^2+b^2\le4\)

Có : a + b + c = 0

=> (a + b)⁵ = (-c)⁵

      a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵ = -c⁵

      a⁵ + b⁵ + c⁵ = -5a⁴b - 10a³b² - 10a²b³ - 5ab⁴

       a⁵ + b⁵ + c⁵ = -5ab(a³ + 2a²b + 2ab² + b³)

      a⁵ + b⁵ + c⁵ = -5ab[(a³ + b³) + (2a²b + 2ab²)]

      a⁵ + b⁵ + c⁵ = -5ab[(a + b)(a² - ab + b²) + 2ab(a + b)]

      a⁵ + b⁵ + c⁵ = -5ab(a + b)(a² + b² + ab)  

      a⁵ + b⁵ + c⁵ = 5abc(a² + b² + ab)   (do a+b+c=0=> a+b=-c)

      2(a⁵ + b⁵ + c⁵) = 5abc(2a² + 2b² + 2ab)

      2(a⁵ + b⁵ + c⁵) = 5abc[a² + b² +(a² + 2ab + b²)]

      2(a⁵ + b⁵ + c⁵) = 5abc[a² + b² + (a + b)²]

      2(a⁵ + b⁵ + c⁵) = 5abc(a² + b² + c²)    (do a+b=-c=> (a +b )² = c²

    \(\Leftrightarrow\) \(a^5+b^5+c^5=\dfrac{5}{2}abc\left(a^2+b^2+c^2\right)\)

Vậy...

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:

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 $a=b=\frac{1}{\sqrt{2}}$