\(\Leftrightarrow\dfrac{2bc}{2bc+a^2}+\dfrac{2ac}{2ac+b^2}+\dfrac{2ab}{2ab+c^2}\le2\)

\(\Leftrightarrow\dfrac{2bc}{2bc+a^2}-1+\dfrac{2ac}{2ac+b^2}-1+\dfrac{2ab}{2ab+c^2}-1\le2-3\)

\(\Leftrightarrow\dfrac{a^2}{2bc+a^2}+\dfrac{b^2}{2ac+b^2}+\dfrac{c^2}{2ab+c^2}\ge1\)

BĐT trên đúng theo C-S:

\(\dfrac{a^2}{2bc+a^2}+\dfrac{b^2}{2ac+b^2}+\dfrac{c^2}{2ab+c^2}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)

Dấu "=" xảy ra khi \(a=b=c\)

Lời giải:

Ta có:
\(\text{VT}=1-\frac{2ab^2}{2ab^2+1}+1-\frac{2bc^2}{2bc^2+1}+1-\frac{2ca^2}{2ca^2+1}\)

\(\text{VT}=3-\underbrace{\left( \frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}\right)}_{N}\) (1)

Áp dụng BĐT Am-Gm:

\(2ab^2+1=ab^2+ab^2+1\geq 3\sqrt[3]{a^2b^4}\)

\(\Rightarrow \frac{2ab^2}{2ab^2+1}\leq \frac{2ab^2}{3\sqrt[3]{a^2b^4}}=\frac{2}{3}\sqrt[3]{ab^2}\)

Tương tự với các phân thức còn lại và cộng theo vế, suy ra :

\(N\leq \frac{2}{3}(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2})\)

Áp dụng BĐT AM-GM:

\(\sqrt[3]{ab^2}\leq \frac{a+b+b}{3}\); \(\sqrt[3]{bc^2}\leq \frac{b+c+c}{3}; \sqrt[3]{ca^2}\leq \frac{c+a+a}{3}\)

\(\Rightarrow N\leq \frac{2}{3}\left(\frac{a+b+b}{3}+\frac{b+c+c}{3}+\frac{c+a+a}{3}\right)\)

\(\Leftrightarrow N\leq \frac{2}{3}(a+b+c)=2\) (2)

Từ \((1),(2)\Rightarrow \text{VT}\geq 1\)

Dấu bằng xảy ra khi \(a=b=c=1\)

Áp dụng BĐT B.C.S ta có :

\(\dfrac{1}{2ab^2+1}+\dfrac{1}{2bc^2+1}+\dfrac{1}{2ca^2+1}\ge\dfrac{9}{2ab^2+2bc^2+2ca^2+3}\)

Ta phải chứng minh \(\dfrac{9}{2ab^2+2bc^2+2ca^2+3}\ge1\)

\(\Leftrightarrow2ab^2+2bc^2+2ac^2+3\le9\) do a,b,c dương nên chia cả hai vế cho abc ta được: \(2\left(a+b+c\right)+\dfrac{3}{abc}\le\dfrac{9}{abc}\)

\(\Leftrightarrow6\le\dfrac{6}{abc}\Leftrightarrow abc\le1\) Bất đẳng thức cuối luôn đúng thật vậy:

áp dụng BĐT AM - GM :

\(\Rightarrow a+b+c\ge3\sqrt[3]{abc}\Leftrightarrow abc\le1\)

\(\Rightarrowđpcm\)

Ta có:

\(\text{VT}=1-\frac{2ab^2}{2ab^2+1}+1-\frac{2bc^2}{2bc^2+1}+1-\frac{2ca^2}{2ca^2+1}\)

\(\text{VT}=3-\underbrace{\left( \frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2} {2ca^2+1}\right)}_{N}\) (1)

Áp dụng BĐT Am-Gm:

\(2ab^2+1=ab^2+ab^2+1\geq 3\sqrt[3]{a^2b^4}\)

\(\Rightarrow \frac{2ab^2}{2ab^2+1}\leq \frac{2ab^2}{3\sqrt[3]{a^2b^4}}=\frac{2}{3}\sqrt[3]{ab^2}\)

Tương tự với các phân thức còn lại và cộng theo vế, suy ra :

\(N\leq \frac{2}{3}(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2})\)

Áp dụng BĐT AM-GM:

\(\sqrt[3]{ab^2}\leq \frac{a+b+b}{3}\); \(\sqrt[3]{bc^2}\leq \frac{b+c+c}{3}; \sqrt[3]{ca^2}\leq \frac{c+a+a}{3}\)

\(\Rightarrow N\leq \frac{2}{3}\left(\frac{a+b+b}{3}+\frac{b+c+c}{3}+\frac{c+a+a}{3}\right)\)

\(\Rightarrow N\leq \frac{2}{3}\left(\frac{a+b+b}{3}+\frac{b+c+c}{3}+\frac{c+a+a}{3}\right)\) (2)

Từ \((1),(2)\Rightarrow \text{VT}\geq 1\)

Dấu bằng xảy ra khi \(a=b=c=1\)

\(A=\dfrac{bc}{8a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}\)

\(=\dfrac{\left(bc\right)^3+8\left(ca\right)^3+8\left(ab\right)^3}{8\left(abc\right)^2}\)

\(=\dfrac{\left(bc\right)^3+\left(2ca\right)^3+\left(2ab\right)^3}{8\left(abc\right)^2}\)

\(=\dfrac{\left(bc\right)^3+\left(2ab+2ca\right)^3-3.2ca.2ab\left(2ab+2ca\right)}{8\left(abc\right)^2}\)

\(=\dfrac{\left(bc\right)^3+\left(-bc\right)^3-3.2ca.2ab.\left(-bc\right)}{8\left(abc\right)^2}\)

\(=\dfrac{12\left(abc\right)^2}{8\left(abc\right)^2}=\dfrac{12}{8}\)

kkkk

Ta có kết quả tổng quát hơn như sau:

Cho $a,b,c \neq 0$ thỏa mãn $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0.$

Chứng minh rằng $$S={\frac {k{a}^{2}-k-1}{{a}^{2}+2\,bc}}+{\frac {{b}^{2}k-k-1}{2\,ac+{b}^{2}}}+{\frac {{c}^{2}k-k-1}{2\,ab+{c}^{2}}}=k$$

Bài 4:

Áp dụng bất đẳng thức Cauchy-shwarz dạng engel ta có:

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

\(=\dfrac{9}{\left(a+b+c\right)^2}=\dfrac{9}{9}=1\)

Dấu " = " xảy ra khi a = b = c = 1

\(\Rightarrowđpcm\)

Bài 1:
Ta có:
\(a^2+b^2-\frac{(a+b)^2}{2}=\frac{2(a^2+b^2)-(a+b)^2}{2}=\frac{(a-b)^2}{2}\geq 0\)

\(\Rightarrow a^2+b^2\geq \frac{(a+b)^2}{2}=\frac{2^2}{2}=2\)

(đpcm)

Dấu "=" xảy ra khi $a=b=1$

Bài 2:

\(P-(ab+bc+ac)=\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{2}\)

\(=\frac{(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)}{2}=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\geq 0, \forall a,b,c\)

\(\Rightarrow P\geq ab+bc+ac\Leftrightarrow P\geq 12\)

Vậy GTNN của $P$ là $12$ khi $a=b=c=2$

\(\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)

\(\Leftrightarrow\dfrac{2}{a+2}-1+\dfrac{2}{b+2}-1+\dfrac{2}{c+2}-1\ge2-3\)

\(\Rightarrow1\ge\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}=\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\)

\(\Rightarrow1\ge\dfrac{\left(a+b+c\right)^2}{a^2+2a+b^2+2b+c^2+2c}\)

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

\(\Rightarrow\) đpcm

Phía trên thoả mãn \(\ge1\) chứ không phải 3/2 đâu ạ