quá đơn giản

ở trên  a(a-b)+b(b-c)+c(c-a)+0 suy ra a=b=c

thay vào k=a^3x3-3a^3=3a^2 -3a+5=3a^2+-3a+5

min của k là min của 3a^2-3a+5 là bằng 17/4

\(P=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\)

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

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

\(=\frac{\frac{1}{a^2}}{\frac{1}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{b}+\frac{1}{a}}\)

Đặt \(\left\{\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\) suy ra \(xyz=1\). Khi đó:

\(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

Áp dụng BĐT AM-GM ta có:

\(\left\{\begin{matrix}\frac{x^2}{y+z}+\frac{y+z}{4}\ge x\\\frac{y^2}{x+z}+\frac{x+z}{4}\ge y\\\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\end{matrix}\right.\).Cộng theo vế ta có:

\(P+\frac{x+y+z}{2}\ge x+y+z\)

\(\Rightarrow P\ge\frac{x+y+z}{2}\ge\frac{3}{2}\left(x+y+z\ge3\sqrt[3]{xyz}=3\right)\)

Hmm...

Ta đánh giá:

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

\(=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\) (Áp dụng BĐT Bunhia)

Tương tự CM được:

\(\frac{b}{b+\sqrt{\left(b+c\right)\left(b+a\right)}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\) ; \(\frac{c}{c+\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Cộng vế 3 BĐT trên lại ta được:

\(Vt\le\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)

Dấu "=" xảy ra khi: \(a=b=c\)

Ko hiểu chỗ nào ib riêng:)

Ta có \( {\displaystyle \displaystyle \sum }cyc\)\(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}=\Sigma_{cyc}\frac{ab}{\left(a+b\right)\sqrt{1-c^2}}\)\(=\Sigma_{cyc}\frac{ab}{\left(a+b\right)\sqrt{\left(a+b+c\right)^2-c^2}}=\Sigma_{cyc}\frac{ab}{\left(a+b\right)\sqrt{a^2+b^2+2\left(ab+bc+ca\right)}}\)

Áp dụng bất đẳng thức AM-GM có \(\hept{\begin{cases}a^2+b^2+2\left(ab+bc+ca\right)\ge2\left(ab+bc\right)+2\left(ab+ca\right)\\a+b\ge2\sqrt{ab}\end{cases}}\)

Do đó ta có \(\Sigma_{cyc}\frac{ab}{\left(a+b\right)\sqrt{a^2+b^2+2\left(ab+bc+ca\right)}}\le\frac{1}{2}\Sigma_{cyc}\sqrt{\frac{ab}{2\left(ab+bc\right)+2\left(ab+ca\right)}}\)

\(\le\frac{1}{4\sqrt{2}}\Sigma_{cyc}\sqrt{\frac{ab}{ab+bc}+\frac{ab}{ab+ca}}\le\frac{1}{4\sqrt{2}}\sqrt{3}\sqrt{\Sigma_{cyc}\left(\frac{ab}{ab+bc}+\frac{ab}{ab+ca}\right)}\)

Đẳng thức xảy ra khi a=b=c=\(\frac{1}{3}\)

@godatakeshidang

Đoạn Đánh giá có thể lm kĩ hơn không:D

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

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

\(\Rightarrow ab.\left(a-b\right)+c.\left(a-b\right).\left(a+b\right)=0\)

\(\Rightarrow\left(ab+ac+bc\right)\left(a-b\right)=0\)

Vậy : \(\left(ab+bc+ca\right)=0\)

\(\Rightarrow\left(ab+bc+ca\right).\left(b-c\right)=0\)

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

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

kết quả là A=\(2a^3\)

a, A = [ -2; 5)

B= ( - \(\infty\); 3 ]

C=(- \(\infty\) ; 4 )

\(\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\)  (1)

Do \(\left(2a+1\right)^2\ge0\)

\(\left(b+3\right)^4\ge0\)

\(\left(5c-6\right)^2\ge0\)

\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)

\(\left(1\right)\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2=0\)

\(\Rightarrow\left(2a+1\right)^2=0;\left(b+3\right)^4=0;\left(5c-6\right)^2=0\)

*) \(\left(2a+1\right)^2=0\)

\(\Rightarrow2a+1=0\)

\(2a=-1\)

\(a=-\dfrac{1}{2}\)

*) \(\left(b+3\right)^4=0\)

\(\Rightarrow b+3=0\)

\(b=-3\)

*) \(\left(5c-6\right)^2=0\)

\(\Rightarrow5c-6=0\)

\(5c=6\)

\(c=\dfrac{6}{5}\)

Vậy \(a=-\dfrac{1}{2};b=-3;c=\dfrac{6}{5}\)