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Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2=1^2=1\)
\(\Rightarrow2\left(a^2+b^2\right)\ge1\Rightarrow a^2+b^2\ge\dfrac{1}{2}\)
Đẳng thức xảy ra khi \(a=b=\dfrac{1}{2}\)
Điều kiện của đề bài : \(a+b=1\)\(\Leftrightarrow\left(a+b\right)^2=1\) \(\Leftrightarrow a^2+2ab+b^2=1\) ( * )
Ta lại có : \(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\) (*)
Cộng (1) (2) lại tao có :
\(a^2+2ab+b^2+a^2-2ab+b^2=a^2+b^2+a^2+b^2+2\left(a^2+b^2\right)\ge1\)
Mà : \(2\left(a^2+b^2\right)\ge1\Leftrightarrow a^2+b^2\ge\dfrac{1}{2}\)
-Áp dụng BĐT Caushy Schwarz ta có:
\(\dfrac{1^2}{a+1}+\dfrac{1^2}{b+1}\ge\dfrac{\left(1+1\right)^2}{a+b+1+1}=\dfrac{4}{3}\)
-Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
Bài 1:
Ta có: a + b - 2c = 0
⇒ a = 2c − b thay vào a2 + b2 + ab - 3c2 = 0 ta có:
(2c − b)2 + b2 + (2c − b).b − 3c2 = 0
⇔ 4c2 − 4bc + b2 + b2 + 2bc − b2 − 3c2 = 0
⇔ b2 − 2bc + c2 = 0
⇔ (b − c)2 = 0
⇔ b − c = 0
⇔ b = c
⇒ a + c − 2c = 0
⇔ a − c = 0
⇔ a = c
⇒ a = b = c
Vậy a = b = c
\(Gt\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow ab+bc+ca=1\)
\(VT=\frac{2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)
\(=\frac{\frac{2}{x}}{\sqrt{\frac{1}{x^2}+1}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{y^2}+1}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{z^2}+1}}\)
\(=\frac{2a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\sqrt{\frac{2a}{\left(a+b\right)}\cdot\frac{2a}{\left(a+c\right)}}+\sqrt{\frac{2b}{\left(b+a\right)}\cdot\frac{b}{2\left(b+c\right)}}\)\(+\sqrt{\frac{2c}{\left(c+a\right)}\cdot\frac{c}{2\left(c+b\right)}}\)
\(\le\frac{\frac{2a}{a+b}+\frac{2a}{a+c}+\frac{2b}{a+b}+\frac{b}{2\left(b+c\right)}+\frac{2c}{c+a}+\frac{c}{2\left(c+b\right)}}{2}=\frac{9}{4}\)
Xét hiệu VT - VP
\(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ab+b^2}+\dfrac{c+a}{ab+c^2}-\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}=\dfrac{a^2+ab-bc-a^2}{a\left(bc+a^2\right)}+\dfrac{b^2+bc-ac-b^2}{b\left(ac+b^2\right)}+\dfrac{c^2+ac-ab-c^2}{c\left(ab+c^2\right)}=\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}\)
Do a,b,c bình đẳng nên giả sử a\(\ge\)b\(\ge\)c, khi đó \(b\left(a-c\right)\)\(\ge\)0, c(b-a)\(\le\)0, a(c-b)\(\le\)0
\(a^3\ge b^3\ge c^3=>abc+a^3\ge abc+b^3\ge abc+c^3\)=>\(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}\le\dfrac{b\left(a-c\right)}{b\left(ac+b^2\right)}\)
=> VT -VP \(\le\) \(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}=\dfrac{ab-ac}{b\left(ac+b^2\right)}+\dfrac{ac-ab}{c\left(ab+c^2\right)}=\dfrac{a\left(b-c\right)}{b\left(ac+b^2\right)}-\dfrac{a\left(b-c\right)}{c\left(ab+c^2\right)}\)
mà \(\dfrac{1}{b\left(ac+b^2\right)}\le\dfrac{1}{c\left(ab+c^2\right)}\) nên VT-VP <0 đpcm
cho ba số dương \(0\le a\le b\le c\le1\) CMR \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le2\)
Vì \(0\le a\le b\le c\le1\) nên:
\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)
Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)
Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)
Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)
Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)
