\(\left\{{}\begin{matrix}\dfrac{a+b}{6}=\dfrac{b+c}{5}\\\dfrac{a+b}{6}=\dfrac{c+a}{7}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=\dfrac{a}{2}\\c=\dfrac{3a}{4}\end{matrix}\right.\)

\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{\dfrac{a^2}{4}+\dfrac{9a^2}{16}-a^2}{2.\dfrac{a}{2}.\dfrac{3a}{4}}=-\dfrac{1}{4}\)

\(cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{a^2+\dfrac{9a^2}{16}-\dfrac{a^2}{4}}{2a.\dfrac{3a}{4}}=\dfrac{7}{8}\)

\(cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{11}{16}\)

\(P=-\dfrac{1}{4}+\dfrac{14}{8}+\dfrac{44}{16}=\dfrac{17}{4}\)

1: Ta có: \(a^2+b^2+c^2\)

\(=\left(a+b+c\right)^2-2\cdot\left(ab+bc+ca\right)\)

\(=5^2-2\cdot174=-323\)

Với x;y dương, ta có BĐT:

\(x^5+y^5\ge x^2y^2\left(x+y\right)\)

Thật vậy, BĐT tương đương:

\(x^5-x^4y+y^5-xy^4\ge0\)

\(\Leftrightarrow x^4\left(x-y\right)-y^4\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (luôn đúng)

Áp dụng:

\(\Rightarrow A\le\dfrac{ab}{a^2b^2\left(a+b\right)+ab}+\dfrac{bc}{b^2c^2\left(b+c\right)+bc}+\dfrac{ca}{c^2a^2\left(c+a\right)+ca}\)

\(A\le\dfrac{1}{ab\left(a+b\right)+1}+\dfrac{1}{bc\left(b+c\right)+1}+\dfrac{1}{ca\left(c+a\right)+1}\)

\(A\le\dfrac{abc}{ab\left(a+b\right)+abc}+\dfrac{abc}{bc\left(b+c\right)+abc}+\dfrac{abc}{ca\left(c+a\right)+abc}=\dfrac{c}{a+b+c}+\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}=1\)

1) ab=2 (I); bc=3 (II); ca=54 (III)

Lấy (I).(II).(III) ⇒ a² . b² . c² = 324 ⇒ abc = ±18

(II) ⇒ a= ±6 ; (I) ⇒ b= ±1/3 ; (II) ⇒ c= ±9

2) ab=5/3 (I); bc=4/5 (II); ca=3/4 (III)

Lấy (I).(II).(III) ⇒ a² . b² . c² = 1 ⇒ abc = ±1

(II) ⇒ a= ±5/4 ; (I) ⇒ b= ±4/3 ; (II) ⇒ c= ±3/5

3) a(a+b+c)= -12 (I)

    b(a+b+c)= 18 (II)

    c(a+b+c)= 30 (III)

Lấy (I)+(II)+(III) ⇒ (a+b+c)² = 36 ⇒ a+b+c = ±6

TH1 : a=6 ⇒ a= -12/6 = -2 ; b= 18/6 = 3 ; c= 30/6 = 5

TH2 : a=-6 ⇒ a= -12/-6 = 2 ; b= 18/-6 = -3 ; c= 30/-6 = -5

a.

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=7\)

\(S=\dfrac{1}{2}AB.AC.sinA=10\sqrt{3}\)

\(\Rightarrow h_a=\dfrac{2S}{BC}=\dfrac{20\sqrt{3}}{7}\)

\(R=\dfrac{BC}{2sinA}=\dfrac{7\sqrt{3}}{3}\)

b.

\(cosA=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}=-\dfrac{11}{34}\)

\(\Rightarrow sinA=\dfrac{3\sqrt{115}}{34}\)

\(S=\dfrac{1}{2}AB.AC.sinA=6\sqrt{115}\)

\(h_a=\dfrac{2S}{BC}=\dfrac{4\sqrt{115}}{7}\)

\(R=\dfrac{BC}{2sinA}=...\)

Với mọi số thực dương a;b;c ta có BĐT:

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

Tương tự, ta có:

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

\(VT\le\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)

Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)

\(VT\le\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\)

Ta lại có: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

\(\Rightarrow VT\le\dfrac{xyz}{xy\left(x+y\right)+xyz}+\dfrac{xyz}{yz\left(y+z\right)+xyz}+\dfrac{xyz}{zx\left(z+x\right)+xyz}=1\)

\(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca}\)

=\(\frac{1}{abc}.\left(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca}\right)\)

=\(\frac{1}{a^5c+b^5c+abc}+\frac{1}{b^5a+c^5a+abc}+\frac{1}{c^5b+a^5b+abc}\)

\(\le\)\(\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\)

Ta có : a³+b³=(a+b)(a²-ab+b²)\(\ge\)ab(a+b) (cosi)

Tương tự ta được:

b³+c³\(\ge bc\left(b+c\right)\)

c³+a³\(\ge ca\left(c+a\right)\)

Như vậy \(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca}\)

\(\le\)\(\frac{1}{ab\left(a+b\right)+abc}+\frac{1}{bc\left(b+c\right)+abc}+\frac{1}{ca\left(c+a\right)+abc}\)

=\(\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)

=\(\frac{1}{a+b+c}.\left(\frac{a+b+c}{ab+bc+ca}\right)=\frac{1}{ab+bc+ca}\le1\)

​mình tò mò muốn biết BĐT trên đẳng thức khi nào nhỉ

Không phải chới đâu BĐT cuối của bạn không bao giờ =1 được

\(\frac{1}{ab+bc+ac}\le\frac{1}{3}\) Đẳng thức khi a=b=c=1

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