Can you prove this inequality? | Moscow Mathematical Olympiad Problem

Given 𝒂, 𝒃 ∈ℝ^+, 𝒂 +𝒃 =𝟏, prove (𝒂+𝟏/𝒂)^𝟐 +(𝒃+𝟏/𝒃)^𝟐 ≥ 𝟐𝟓/𝟐. How to prove this inequality? We apply the GM-AM inequality to solve this problem.

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A geometric proof: The LHS of the inequality reminds us the squared distance between point P(a, b) and point Q(-1/a, -1/b).
Clearly point P lies on line segment y=-x+1 in the 1st quadrant.
For point Q, letting x=-1/a and solving y=-1/b=-1/(1-a) gives a hyperbolic curve for Q, that is y = -x / (1+x) in the 3rd quadrant for x< -1.
At point (-2, -2), the line with slope -1 is tangent to the curve of Q. Call this tangent line as T: y=-x-4.
Because of concavity, all points on the Q curve lie to the southwest of this tangent line T, except at point (-2, -2).
The distance between line P and line T is √(25/2).
Now consider the distance between P and Q. By using "hypotenuse > any leg" in right ◺s (one to the SW of line T and one to the NE of line T):
PQ distance
≥ (distance between P's line and tangent line T) + (point Q's perpendicular distance to the tangent line T)
= √(25/2) + (distance between point Q and tangent line T).
with equality iff line PQ is perpendicular to line T, i.e., when a=b=1/2. This finishes the proof.