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 1:

Dự đoán dấu "=" xảy ra khi \(a=b=c=1\) ta tính được giá trị là \(9\)

Ta sẽ chứng minh nó là GTLN

Thật vậy ta cần chứng minh

\(\Sigma\dfrac{11a+4b}{4a^2-ab+2b^2}\le\dfrac{3\left(ab+ac+bc\right)}{abc}\)

\(\LeftrightarrowΣ\left(\dfrac{3}{a}-\dfrac{11a+4b}{4a^2-ab+2b^2}\right)\ge0\)

\(\LeftrightarrowΣ\dfrac{\left(a-b\right)\left(a-6b\right)}{a\left(4a^2-ab+2b^2\right)}\ge0\)

\(\LeftrightarrowΣ\left(\dfrac{\left(a-b\right)\left(a-6b\right)}{a\left(4a^2-ab+2b^2\right)}+\dfrac{1}{b}-\dfrac{1}{a}\right)\ge0\)

\(\LeftrightarrowΣ\dfrac{\left(a-b\right)^2\left(a+b\right)}{ab\left(4a^2-ab+2b^2\right)}\ge0\) (luôn đúng)

Bài 2:

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(a^5+b^2+c^2\right)\left(\dfrac{1}{a}+b^2+c^2\right)\ge\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow\dfrac{1}{a^5+b^2+c^2}\le\dfrac{\dfrac{1}{a}+b^2+c^2}{\left(a^2+b^2+c^2\right)^2}\)

Tương tự rồi cộng theo vế ta có:

\(Σ\dfrac{1}{a^5+b^2+c^2}\le\dfrac{Σ\dfrac{1}{a}+2Σa^2}{\left(a^2+b^2+c^2\right)^2}\)

Ta chứng minh \(Σ\dfrac{1}{a}+2\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\) - BĐT cuối đúng

Vậy ta có ĐPCM. Dấu "=" xảy ra khi \(a=b=c=1\)

Bài 3:

Từ \(a+b+c=3abc\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=3\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow xy+yz+xz=3\) và BĐT cần chứng minh là

\(x^3+y^3+z^3\ge3\). Áp dụng BĐT AM-GM ta có:

\(x^3+x^3+1\ge3\sqrt[3]{x^3\cdot x^3\cdot1}=3x^2\)

Tương tự có: \(y^3+y^3+1\ge3y^2;z^3+z^3+1\ge3z^2\)

Cộng theo vế 3 BĐT trên ta có:

\(2\left(x^3+y^3+z^3\right)+3\ge3\left(x^2+y^2+z^2\right)\)

Lại có BĐT quen thuộc \(x^2+y^2+z^2\ge xy+yz+xz\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge3\left(xy+yz+xz\right)=9\left(xy+yz+xz=3\right)\)

\(\Rightarrow2\left(x^3+y^3+z^3\right)+3\ge9\Rightarrow2\left(x^3+y^3+z^3\right)\ge6\)

\(\Rightarrow x^3+y^3+z^3\ge3\). BĐT cuối đúng nên ta có ĐPCM

Đẳng thức xảy ra khi \(a=b=c=1\)

T/b:Vâng, rất giỏi

lần sau đăng từng câu 1 dc ko bn :)

system errow

BĐT cần chứng minh tương đương với

\(\left(1-\frac{a^5-a^2}{a^5+b^2+c^2}\right)+\left(1-\frac{b^5-b^2}{b^5+c^2+a^2}\right)+\left(1-\frac{c^5-c^2}{c^5+a^2+b^2}\right)\le3\)

hay \(\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)

Từ \(abc\ge1\) ta có:

\(\frac{1}{a^5+b^2+c^2}\le\frac{1}{\frac{a^5}{abc}+b^2+c^2}=\frac{1}{\frac{a^4}{bc}+b^2+c^2}\)

\(\le\frac{1}{\frac{2a^4}{b^2+c^2}+b^2+c^2}=\frac{b^2+c^2}{2a^4+\left(b^2+c^2\right)^2}\)

Do \(4u^2+v^2\ge4uv\Leftrightarrow4u^2+v^2\ge\frac{2}{3}\left(u+v\right)^2\)nên 

\(2a^4+\left(b^2+c^2\right)^2\ge\frac{2}{3}\left(a^2+b^2+c^2\right)^2\)

Suy ra \(\frac{1}{a^5+b^2+c^2}\le\frac{3\left(b^2+c^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

Tương tự ta có \(\frac{1}{b^5+c^2+a^2}\le\frac{3\left(c^2+a^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

và \(\frac{1}{c^5+a^2+b^2}\le\frac{3\left(a^2+b^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

Cộng ba vế của các BĐT trên ta được

\(\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)

Vậy \(\frac{a^5-a^2}{a^5+b^2+c^2}+\frac{b^5-b^2}{b^5+c^2+a^2}+\frac{c^5-c^2}{c^5+a^2+b^2}\ge0\)

(Dấu "="\(\Leftrightarrow a=b=c=1\))

BĐT bị ngược dấu, BĐT đúng phải là:

\(\dfrac{a}{ac+4}+\dfrac{b}{ab+4}+\dfrac{c}{bc+4}\le\dfrac{a^2+b^2+c^2}{16}\)

Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)

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

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

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

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

\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)

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

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

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

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

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

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

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

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)