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

\(\Rightarrow A^2=\left(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\right)^2\)\(=\left(\sqrt{c}.\sqrt{a-c}+\sqrt{c}.\sqrt{b-c}\right)^2\)

\(\Rightarrow A^2\le\left(c+b-c\right)\left(c+ a-c\right)\left(\text{ Bunhiacopxki}\right)\)

\(\Rightarrow A^2\le ab\Leftrightarrow A\le\sqrt{ab}\left(đpcm\right)\)

\(\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{\left(a+c\right)\left(b+c\right)}+\sqrt{\left(a-c\right)\left(b-c\right)}\right)^2\)

\(\le\left(a+c+a-c\right)\left(b+c+b-c\right)\)

\(=2a\cdot2b=4ab=VP^2\)

\(\Rightarrow VT\le VP\) *ĐPCM*

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\([\sqrt{c(a-c)}+\sqrt{c(b-c)}]^2\leq [c+(b-c)][(a-c)+c]=ab\)

\(\Rightarrow \sqrt{c(a-c)}+\sqrt{c(b-c)}\leq \sqrt{ab}\) (đpcm)

Dấu "=" xảy ra khi $a=b=2c$

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)

\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)

Từ ( 1 ) và ( 2 ) có đpcm

Xét \(\sqrt{\dfrac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\dfrac{\left(a\left(a+b+c\right)+bc\right)\left(b\left(a+b+c\right)+ac\right)}{c\left(a+b+c\right)+ab}}\)

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

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

Tương tự cho 2 đẳng thức còn lại rồi cộng theo vế

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

Đề đánh bị lỗi.

Áp dụng bất đẳng thức Bunhiacopski:

\(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\le\sqrt{\left[\sqrt{c}^2+\sqrt{\left(a-c\right)}^2\right]\left[\sqrt{c}^2+\sqrt{\left(b-c\right)}^2\right]}\)

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

Đặt \(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\)  = A

Ta có A^2 = \(\left(\sqrt{\left(a-c\right).c}+\sqrt{c.\left(b-c\right)}\right)^2\)

Áp dụng bđt bunhiacopxki ta có A^2 <= \(\left(\sqrt{a-c}^2+\sqrt{c^2}\right).\left(\sqrt{c^2}+\sqrt{b-c^2}\right)\)

                                                       = (a-c+c).(c+b-c) = ab

<=> A<= \(\sqrt{ab}\)=> ĐPCM

Dấu"=" <=> a-c = c và c = b-c

<=> a=b=2c>0

Ta có bất đẳng thức bunhicopxki

\(\sqrt{ax}+\sqrt{by}\le\sqrt{\left(a+x\right)\left(b+y\right)}\)

Áp dụng vào ta có:

\(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{\left(a-c+c\right)\left(b-c+c\right)}\le\sqrt{ab}\)

Dấu bằng xảy ra khi a-c = b-c

Đặt: \(\frac{c}{a}=x;\frac{c}{b}=y\left(0< x;y< 1\right)\)

\(BĐT\Leftrightarrow\sqrt{\frac{c}{b}\left(1-\frac{c}{a}\right)}+\sqrt{\frac{c}{a}\left(1-\frac{c}{b}\right)}\le1\)

\(\Leftrightarrow\sqrt{y\left(1-x\right)}+\sqrt{x\left(1-y\right)}\le1\)

Áp dụng BĐT AM-GM, ta có:

\(\sqrt{y\left(1-x\right)}\le\frac{y+1-x}{2}\)

\(\sqrt{x\left(1-y\right)}\le\frac{x+1-y}{2}\)

\(\RightarrowĐPCM\)

\(\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\frac{\left(a^2+ab+ac+bc\right)\left(b^2+bc+ba+ac\right)}{c^2+ca+cb+ab}}=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}=a+b\left(a,b,c>0;a+b+c=1\right)\)

Bạn làm tương tự nha

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