Đề này sai đó bạn.

Giả sử c = 2,5; a = 2 và c = 1,5

Ta có: \(c\ge a;c\ge b\) nhưng \(c< a+b\) (mâu thuẫn với đề bài).

\(a\sqrt{b-1}+b\sqrt{a-1}-1\)

\(=a\sqrt{1.\left(b-1\right)}+b\sqrt{1.\left(a-1\right)}\le a\dfrac{1+b-1}{2}+b\dfrac{1+a-1}{2}=\dfrac{ab}{2}+\dfrac{ab}{2}=ab\)dấu "=" xảy ra khi a=b=2

1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

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

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

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

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)

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

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

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

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

b)Đã cm

c)\(a^2+b^2+1\ge ab+a+b\)

\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)

Dấu bằng xảy ra khi a=b=1

2. a) a² + \(\dfrac{b^2}{4}\)≥ab

<=> a² - ab + \(\dfrac{b^2}{4}\)≥ 0

<=> a² -2.\(\dfrac{b}{2}a+\left(\dfrac{b}{2}\right)^2\) ≥ 0

<=> \(\left(a-\dfrac{b}{2}\right)^2\)≥ 0 ( luôn đúng )

=> đpcm

b) ( a + b)² ≤ 2( a² + b²)

<=> a² + 2ab + b² - 2a² - 2b² ≤ 0

<=> - ( a² - 2ab + b² ) ≤ 0

<=> - ( a - b)² ≤ 0 ( luôn đúng )

=> đpcm

c) a² + b² + 1 ≥ ab + a + b

<=> 2( a² + b² + 1 ) ≥ 2( ab + a + b)

<=> a² - 2ab + b² + a² - 2a + 1 + b² - 2b + 1 ≥ 0

<=> ( a - b)² + ( a - 1)² + ( b - 1)² ≥ 0 ( luôn đúng )

=> đpcm

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

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

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

\(\Rightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (đúng)

Dấu "=" xảy ra khi: \(a=b\)

ta có :

\(\frac{ax+by}{2}\ge\frac{a+b}{2}.\frac{x+y}{2}\Leftrightarrow2\left(ax+by\right)\ge\left(a+b\right)\left(x+y\right)\)

\(\Leftrightarrow2\left(ax+by\right)\ge ax+ay+bx+by\)

\(\Leftrightarrow ax-ay+by-bx\ge0\Leftrightarrow\left(a-b\right)\left(x-y\right)\ge0\)

Điều này đúng do giả thuyết \(a\ge b,x\ge y\)

ta có \(\dfrac{ax+by}{2}\) ≥ \(\dfrac{a+b}{2}\). \(\dfrac{x+y}{2}\)

<=> 2(ax + by) ≥ (a + b)(x + y)

<=> 2(ax + by) ≥ ax + ay + bx + by

<=> ax + by - ay - bx ≥ 0

<=> (a - b)(x - y) ≥ 0 (luôn đúng vì giả thiết a ≥ b và x ≥ y)

vậy nếu a ≥ b, x ≥ y thì \(\dfrac{ax+by}{2}\) ≥ \(\dfrac{a+b}{2}\). \(\dfrac{x+y}{2}\)

Ta có \dfrac{ax + by}2 \ge \dfrac{a + b}2 . \dfrac{x + y}22ax+by​≥ 2a+b​. 2x+ y​

$\iff 2(ax+by)\ge (a+b)(x+y)$

$\iff 2(ax+by)\ge ax+ay+bx+by$

$\iff ax+by-ay-bx\ge 0$

$\iff (a-b)(x-y)\ge 0$ (luôn đúng vì giả thiết $a\ge b$ và $x\ge y$).

Vậy nếu $a\ge b$, $x\ge y$ thì \dfrac{ax + by}2 \ge \dfrac{a + b}2 . \dfrac{x + y}22ax+by​≥ 2a+b​. 2x+ y​.

bạn thử tải ảnh về xem nhé

1) xét hiệu

\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}\ge0\)

<=> \(\dfrac{b\left(a+b\right)}{ab\left(a+b\right)}+\dfrac{a\left(a+b\right)}{ab\left(a+b\right)}-\dfrac{4ab}{ab\left(a+b\right)}\ge0\)

=> b(a+b)+a(a+b)-4ab ≥ 0

<=> ab+b²+a²+ab-4ab ≥ 0

<=> a² -2ab+b² ≥ 0

<=> (a-b)² ≥ 0 (luôn đúng )

=> đpcm

2)Ta có:\(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

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

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

TT\(\Rightarrow\left(b+c\right)^2\ge4bc;\left(c+a\right)^2\ge4ca\)

\(\Rightarrow\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\ge64a^2b^2c^2\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)

Áp dụng BĐT cô si cho 2 số không âm

\(b+c\ge2\sqrt{bc}\)

<=>\(\left(b+c\right)^2\ge4bc\) (1)

Áp dụng BĐT cô si cho 2 số không âm

\(a+\left(b+c\right)\ge2\sqrt{a\left(b+c\right)}\)

<=>\(\left[a+\left(b+c\right)\right]^2\ge4a\left(b+c\right)\)

<=>\(1\ge4a\left(b+c\right)\) (2)

nhân (1) với (2) ta đc

\(\left(b+c\right)^2\ge16abc.\left(b+c\right)\)

<=>\(b+c\ge16abc\) (đpcm)

\(1=\left(a+b+c\right)^2\ge4a\left(b+c\right)\)

\(\Rightarrow b+c\ge4a\left(b+c\right)^2\ge4a\cdot4bc=16abc\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=c\\a+b+c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{2}\\b=c=\frac{1}{4}\end{matrix}\right.\)

Sửa: Cho a+b<0

\(BĐT\Leftrightarrow\dfrac{\left(a+b\right)^3}{8}\ge\dfrac{a^3+b^3}{2}\\ \Leftrightarrow2\left(a+b\right)^3\ge8\left(a^3+b^3\right)\\ \Leftrightarrow2\left(a^3+b^3\right)+6ab\left(a+b\right)\ge8\left(a^3+b^3\right)\\ \Leftrightarrow6ab\left(a+b\right)-6\left(a^3+b^3\right)\ge0\\ \Leftrightarrow6\left[ab\left(a+b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\right]\ge0\\ \Leftrightarrow6\left(a+b\right)\left(-a^2+2ab-b^2\right)\ge0\\ \Leftrightarrow-6\left(a+b\right)\left(a-b\right)^2\ge0\left(\text{luôn đúng do }-6< 0;a+b< 0\right)\)

Dấu \("="\Leftrightarrow a=b< 0\)