\(\Leftrightarrow a^4-a^3b+b^4-ab^3>=0\)

\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)>=0\)

\(\Leftrightarrow\left(a-b\right)^2\cdot\left(a^2+ab+b^2\right)>=0\)(luôn đúng)

ta có:\(\left(\sqrt{a}-\sqrt{b}\right)\ge0\)

\(\Rightarrow a-2\sqrt{ab}+b\ge0\)

\(\Rightarrow a+b\ge2\sqrt{ab}\)

\(\Rightarrow\frac{a+b}{2}\ge\sqrt{ab}\)

dấu "=" xảy ra khi a=b

Cho  A bằng 34x89y

tìm x y biết:

A chia hết cho 4 chia hết cho 3 chia 2 dư1 chia 5 dư 4

tích đúng cho ai hợp lý

<=>  \(a+b\ge2\sqrt{ab}\)

<=> \(a+b-2\sqrt{ab}\ge0\)

<=. \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(luôn đúng )

dấu = khi a=b

\(VT=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

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

Mà theo Cô-si ta có:

\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) (hằng đẳng thức)

\(\Rightarrow VT\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi a=b=c

Đặt b + c = x ; c + a = y ; a + b = z
=> a = (y + z - x) / 2 ; b = (x + z - y) / 2 ; c = (x + y - z) / 2
=> P = a/b+c + b/c+a + c/a+b = (y + z - x) / 2x + (x + z - y) / 2y + (x + y - z) / 2z
= 1/2. (y/x + z/x - 1 + x/y + z/y - 1 + x/z + y/z - 1) = 1/2. (x/y + y/x + x/z + z/x + y/z + z/y - 3)

Áp dụng BĐT A/B + B/A ≥ 0 hoặc Cô-si cũng được
=> P ≥ 1/2. (2 + 2 + 2 - 3) = 3/2 (đpcm)

Dấu = xảy ra <=> x = y = z <=> b+c = c+a = a+b <=> a = b = c

Ta có \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\Leftrightarrow a+b-2\sqrt{ab}\ge0\Leftrightarrow\frac{a+b}{2}\ge\sqrt{ab}.\)

\(\frac{a^2+2ab+b^2}{4}\ge ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\)

vì BĐT cuối đúng nên BĐT đầu đúng

Cảm ơn bạn nhiều nha....

\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

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

Áp dụng BĐT Cauchy-Schwarz ta có:

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

Ta c/m BĐT phụ: \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\)( b tự c/m nhé. Chuyển vế, c/m VP>=0 là xong )

\(\Rightarrow\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{2.\frac{1}{3}\left(a+b+c\right)^2}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)

                                               đpcm

Có thể c/m luôn giùm bđt phụ không ạ?

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

\(\Leftrightarrow a^2-2\cdot\dfrac{1}{2}\cdot a\cdot b+\left(\dfrac{1}{2}b\right)^2\ge0\)

\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)

Áp dụng bđt cô-si có:

\(a^2+\dfrac{b^2}{4}\ge2\sqrt{\dfrac{a^2b^2}{4}}=2\cdot\dfrac{ab}{2}=ab\left(đpcm\right)\)